Problem Description

There are two strings A and B with equal length. Both strings are made up of lower case letters. Now you have a powerful string painter. With the help of the painter, you can change a segment of characters of a string to any other character you want. That is, after using the painter, the segment is made up of only one kind of character. Now your task is to change A to B using string painter. What’s the minimum number of operations?

 

Input

Input contains multiple cases. Each case consists of two lines:
The first line contains string A.
The second line contains string B.
The length of both strings will not be greater than 100.

 

Output

A single line contains one integer representing the answer.

 

Sample Input

 

Sample Output

#include<iostream>

#include<cstdio>

#include<algorithm>

#include<cstring>

using namespace std;

#define ll long long

const int maxn=1e5+;

const int INF=0x3f3f3f3f;

char str1[],str2[];

int dp[][];

int ans[];

int main()

{

    //freopen("in.txt","r",stdin);

    while(cin>>str1+>>str2+)

    {

        int n=strlen(str1+);

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

        for(int d=;d<n;d++)

        for(int i=;i+d<=n;i++){

            int j=i+d;

            dp[i][j]=dp[i+][j]+;

            for(int k=i+;k<=j;k++)

                if(str2[i]==str2[k])

                    dp[i][j]=min(dp[i][j],dp[i+][k]+dp[k+][j]);

        }

        for(int i=; i<=n; i++)ans[i]=dp[][i];

        for(int i=; i<=n; i++)

        if(str1[i]==str2[i]) ans[i]=ans[i-];

        else  for(int j=; j<i; j++) ans[i]=min(ans[i],ans[j]+dp[j+][i]);

        printf("%d\n",ans[n]);

    }

    return ;

}