Gym - 100989E

Islam is usually in a hurry. He often types his passwords incorrectly. He hates retyping his password several times whenever he tries to login, especially that his passwords are usually very long. He believes that websites should be tolerant with very long passwords. In other words, he believes that if a password is very long, and there is only one mistake in the password, the website should allow the user to login.

Your task is to check if an entered password should be accepted according to Islam, or not. The entered password will be accepted if it matches the user’s password, or if the user’s password length is at least 8 characters and the user made a mistake with only one character (either replaced it with a wrong character or dropped it).

Given the user’s password, and the entered password, determine if the entered password should be accepted according to Islam.

Input

The first line of input contains the user’s password.

The second line of input contains the entered password.

Both strings contain only lowercase and uppercase English letters.

The length of each string is at least 1 and at most 100.

Output

Print yes if the entered password should be accepted according to Islam, otherwise print no.

Examples

Input

AgentMahone
IslamIsMahone

Output

no

Input

ofmahone
ofmahome

Output

yes

Input

algorithms
algorthms

Output

yes

Input

Mahone
mahonE

Output

no

题意：输入两个字符串，如果第一个字符串的位数为8位以下，则两个字符串必须完全相同才输出“yes”，否则输出“no”。当第一个字符串的长度大于8位时，则第二个字符串可以比第一个字符串少一个字符，或与第一个字符串一个字符不相同，输出即为“yes”，其余情况均输出“no”。
题解：可分为四种情形，我们可以对其逐一进行判断，第一种：是第一个字符串比第二个字符串多两个字符，或者第二个字符串比第一个字符串多，即输出“no”；第二种，第一个字符串长度小于8，两个字符串必须完全相同才“yes”；第三种，两个字符串长度相等，定义一个变量flag，对其不相同的次数进行计数，超过2即为“no”；第四种就是第一个字符串比第二个字符串多1个字符的情形。注意：各种情形的判断顺序不可随意调动。具体情况详见代码所示：

#include<iostream> 
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<stack>
#include<queue>
#define ll long long
using namespace std;
int main(){
    string s1,s2;
    cin>>s1>>s2;
    int flag=0;
    int a=s1.length(),b=s2.length();
    if(a<b||a-b>1) cout<<"no"<<endl;
    else if(a<8)
    {
        if(s1==s2) cout<<"yes"<<endl;
        else cout<<"no"<<endl;
    }
    else if(a==b)
    {
        for(int i=0;i<a;i++)
        {
            if(s1[i]!=s2[i]) flag++;
        }
        if(flag>1) cout<<"no"<<endl;
        else cout<<"yes"<<endl;
    }
    else
    {
        int num1=0,num2=0;
        while(num1<a&&num2<b)
        {
            if(s1[num1]!=s2[num2])
            {  //当第一个字符串比第二个多一个字符时，不一样时，只增加第一个字符串的下标 
                num1++;
                flag++;
            }
            else
            {
                num1++;
                num2++;
            }
            if(flag==2) break;
        }
        if(flag==2) cout<<"no"<<endl;
        else cout<<"yes"<<endl;
    }
    return 0;
}