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.

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

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

#include<cstring>
#include<cstdio>
#include<string>
#include<iostream>
#include<algorithm>
#define max_v 1005
using namespace std;
char x[max_v],y[max_v];
int dp[max_v][max_v];
int l1,l2;
int main()
{while(~scanf("%d %s",&l1,x)){scanf("%d %s",&l2,y);memset(dp,0,sizeof(dp));for(int i=1; i<=l1; i++){for(int j=1; j<=l2; j++){if(x[i-1]==y[j-1]){dp[i][j]=dp[i-1][j-1]+1;}else{dp[i][j]=max(dp[i-1][j],dp[i][j-1]);}}}int t=l1;if(l2>l1)t=l2;printf("%d\n",t-dp[l1][l2]);}return 0;
}