Given two sequences of numbers : a[1], a[2], …… , a[N], and b[1], b[2], …… , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], …… , a[K + M – 1] = b[M]. If there are more than one K exist, output the smallest one.
Input
The first line of input is a number T which indicate the number of cases. Each case contains three lines. The first line is two numbers N and M (1 <= M <= 10000, 1 <= N <= 1000000). The second line contains N integers which indicate a[1], a[2], …… , a[N]. The third line contains M integers which indicate b[1], b[2], …… , b[M]. All integers are in the range of [-1000000, 1000000].
Output
For each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead.
Sample Input

2
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 1 3
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 2 1

Sample Output

6
-1

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;

int ne[10010];
int a[1000010], b[10010];

void GetNext(int lenb)
{
    int i = 0, j = -1;
    ne[0] = -1;
    while(i < lenb)
    {
        if(j == -1 || b[i] == b[j]){
            ne[++i] = ++j;
        }
        else j = ne[j];
    }
}

int KmpCompare(int lena, int lenb)
{
    int i = 0, j = 0;
    while(i < lena && j < lenb)
    {
        if(j == -1 || a[i] == b[j]){
            i++;
            j++;
            if(j == lenb) return i - j + 1;
        }
        else j = ne[j];
    }
    return -1;
}

int main()
{
    int t, c, d, e, i;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d %d", &c, &d);
        for(i=0; i<c; i++){
            scanf("%d", &a[i]);
        }
        for(i=0; i<d; i++){
            scanf("%d", &b[i]);
        }
        GetNext(d);
        e = KmpCompare(c, d);
        printf("%d\n", e);
    }
    return 0;
}