AGTC

Description

Let x and y be two strings over some finite alphabet A. We would like to transform x into y allowing only operations given below:

• Deletion: a letter in x is missing in y at a corresponding position.
• Insertion: a letter in y is missing in x at a corresponding position.
• Change: letters at corresponding positions are distinct

Certainly, we would like to minimize the number of all possible operations.

Illustration

A G T A A G T * A G G C

| | |       |   |   | |

A G T * C * T G A C G C

Deletion: * in the bottom line

Insertion: * in the top line

Change: when the letters at the top and bottom are distinct

This tells us that to transform x = AGTCTGACGC into y = AGTAAGTAGGC we would be required to perform 5 operations (2 changes, 2 deletions and 1 insertion). If we want to minimize the number operations, we should do it like

A  G  T  A  A  G  T  A  G  G  C

|  |  |        |     |     |  |

A  G  T  C  T  G  *  A  C  G  C

and 4 moves would be required (3 changes and 1 deletion).

In this problem we would always consider strings x and y to be fixed, such that the number of letters in x is m and the number of letters in y is n where n ≥ m.

Assign 1 as the cost of an operation performed. Otherwise, assign 0 if there is no operation performed.

Write a program that would minimize the number of possible operations to transform any string x into a string y.

Input

The input consists of the strings x and y prefixed by their respective lengths, which are within 1000.

Output

An integer representing the minimum number of possible operations to transform any string x into a string y.

Sample Input

10 AGTCTGACGC
11 AGTAAGTAGGC

Sample Output

4

#include <stdio.h>
char s1[1001], s2[1001];
int dp[1001];
int main()
{
    int lenS1, lenS2, old, t;
    while(1)
    {
        if(EOF==scanf("%d %s", &lenS1, s1))break;
        scanf("%d %s", &lenS2, s2);

        for(int i=0; i<=lenS2; i++)
            dp[i]=i;

        for(int i=0; i<lenS1; i++)
        {
            old = dp[0];
            dp[0] = i+1;
            for(int j=0; j<lenS2; j++)
            {
                t=dp[j+1];
                if(s1[i]==s2[j])
                {
                    //dp[j+1]=dp[j];
                    dp[j+1]=old;
                }
                else
                {
                    if(dp[j+1]>dp[j])dp[j+1]=dp[j];
                    if(dp[j+1]>old)dp[j+1]=old;
                    ++dp[j+1];
                }
                old=t;
            }
        }
        printf("%d\n", dp[lenS2]);
    }
    return 0;
}