Prime Path

Description

The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-digit room numbers on their offices. 

— It is a matter of security to change such things every now and then, to keep the enemy in the dark. 

— But look, I have chosen my number 1033 for good reasons. I am the Prime minister, you know! 

— I know, so therefore your new number 8179 is also a prime. You will just have to paste four new digits over the four old ones on your office door. 

— No, it’s not that simple. Suppose that I change the first digit to an 8, then the number will read 8033 which is not a prime! 

— I see, being the prime minister you cannot stand having a non-prime number on your door even for a few seconds. 

— Correct! So I must invent a scheme for going from 1033 to 8179 by a path of prime numbers where only one digit is changed from one prime to the next prime. 

Now, the minister of finance, who had been eavesdropping, intervened. 

— No unnecessary expenditure, please! I happen to know that the price of a digit is one pound. 

— Hmm, in that case I need a computer program to minimize the cost. You don’t know some very cheap software gurus, do you? 

— In fact, I do. You see, there is this programming contest going on… Help the prime minister to find the cheapest prime path between any two given four-digit primes! The first digit must be nonzero, of course. Here is a solution in the case above. 

1033

1733

3733

3739

3779

8779

8179

The cost of this solution is 6 pounds. Note that the digit 1 which got pasted over in step 2 can not be reused in the last step – a new 1 must be purchased.

Input

One line with a positive number: the number of test cases (at most 100). Then for each test case, one line with two numbers separated by a blank. Both numbers are four-digit primes (without leading zeros).

Output

One line for each case, either with a number stating the minimal cost or containing the word Impossible.

Sample Input

3
1033 8179
1373 8017
1033 1033

Sample Output

6
7
0

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 100001
const int inf=0x7fffffff;   //无限大
const int MAXN = 10000;
bool flag[MAXN];
int primes[MAXN], pi;
struct point
{
    int x;
    int y;
};
void GetPrime_1()
{
    int i, j;
    pi = 0;
    memset(flag, false, sizeof(flag));
    for (i = 2; i < MAXN; i++)
        if (!flag[i])
        {
            primes[i] = 1;//素数标识为1
            for (j = i; j < MAXN; j += i)
                flag[j] = true;
        }
}
int vis[maxn];
int main()
{
    GetPrime_1();
    int t;
    cin>>t;
    while(t--)
    {
        memset(vis,0,sizeof(vis));
        int n,m;
        cin>>n>>m;
        vis[n]=1;
        queue<point> q;
        q.push((point){n,0});
        int flag1=0;
        while(!q.empty())
        {
            point now=q.front();
            if(now.x==m)
            {
                flag1=now.y;
                break;
            }
            point next;
            for(int i=0;i<=9;i++)
            {
                next.x=now.x/10;
                next.x*=10;
                next.x+=i;
                next.y=now.y+1;
                if(next.x<1000||next.x>=10000)
                    continue;
                if(vis[next.x]==1)
                    continue;
                if(next.x==m)
                {
                    flag1=next.y;
                    break;
                }
                if(primes[next.x]==1)
                {
                    //cout<<next.x<<endl;
                    vis[next.x]=1;
                    q.push(next);
                }

            }
            for(int i=0;i<=9;i++)
            {
                int temp=now.x%10;
                next.x=now.x/100;
                next.x*=100;
                next.x+=i*10;
                next.x+=temp;
                if(next.x<1000||next.x>=10000)
                    continue;
                if(vis[next.x]==1)
                    continue;
                if(next.x==m)
                {
                    flag1=next.y;
                    break;
                }
                if(primes[next.x]==1)
                {
                    //cout<<next.x<<endl;
                    vis[next.x]=1;
                    q.push((point){next.x,now.y+1});
                }
            }
            for(int i=0;i<=9;i++)
            {
                int temp=now.x%100;
                next.x=now.x/1000;
                next.x*=1000;
                next.x+=i*100;
                next.x+=temp;
                if(next.x<1000||next.x>=10000)
                    continue;
                if(vis[next.x]==1)
                    continue;
                if(next.x==m)
                {
                    flag1=next.y;
                    break;
                }
                if(primes[next.x]==1)
                {
                    //cout<<next.x<<endl;
                    vis[next.x]=1;
                    q.push((point){next.x,now.y+1});
                }
            }
            for(int i=0;i<=9;i++)
            {
                int temp=now.x%1000;
                next.x=now.x/10000;
                next.x*=10000;
                next.x+=i*1000;
                next.x+=temp;
                if(next.x<1000||next.x>=10000)
                    continue;
                if(vis[next.x]==1)
                    continue;
                if(next.x==m)
                {
                    flag1=next.y;
                    break;
                }
                if(primes[next.x]==1)
                {
                //    cout<<next.x<<endl;
                    vis[next.x]=1;
                    q.push((point){next.x,now.y+1});
                }
            }
            if(flag1>0)
            break;
            q.pop();
        }
    printf("%d\n",flag1);
    }
    return 0;
}