I. Same String

There are two lengths of only lowercase letters to n strings S . 1 , S 2 , and m groups of letters corresponding relationship, the relationship between each group of two-letter c . 1 and c 2 , representing c . 1 can directly into c 2 , you need to determine S . 1 whether by this m is converted to a group relationship S 2 .

Input Format

A first line of input n- ( . 1 ≤ n- ≤ 100 ) , representative length of the string.
The second and third rows two input strings S . 1 , S 2 .
A fourth line input m ( . 1 ≤ m ≤ 325 ) , there are representative of m groups relationship.
Subsequently m -th row, the i row two characters U i , V i , the representative U i can be directly changed to V i .

Output Format

If S . 1 by the m conversion relationship becomes set S 2 , output "YES", and otherwise outputs "NO".

Sample

input

6
aabbcc
cdbcad
4
a c
c a
a d
b c

output

YES

prompt

Can be converted many times, such can be converted to a b, and b may be converted is c, then a can be converted to c.
Sample a: aabbcc-> cabbcc-> cdbbcc-> cdbccc- > cdbcac-> cdbcaa-> cdbcad

Thinking: determining character by character, each character is judged whether s1 s2 corresponding character can be achieved

#include<iostream>
#include<string>
#include<vector>
#include<queue> 
#include<cstring>
using namespace std;
int n;
string s1,s2;
int mark[500];
char tt;
int flag=0;
vector<char>ve[500];
void bfs(char x){
    queue<char >que;
    que.push(x);
    int c=int (x);
    mark[c]=1;
    while(que.size()){
        char y=que.front();
        que.pop();
        if(y==tt){
            flag=1;
            break;
        }
        char yy;
        int yyy;
        for(int i=0;i<ve[y].size();i++){
            yy=ve[y][i];
            yyy=int(yy);
            if(mark[yyy]==0&&yy!=y){
                mark[yyy]=1;
                que.push(yyy);
            }
        }
    }
}
int main(){
    cin>>n;
    cin>>s1>>s2;
    int t;
    cin>>t;
    char a,b;
    for(int i=0;i<t;i++){
        cin>>a>>b;
        ve[a].push_back(b);//表示a可以变成b
    }
    for(int i=0;i<n;i++)
    {
        if(s1[i]==s2[i])
            continue ;
        else {
            tt=s2[i];
            memset(mark,0,sizeof(mark));
            flag=0;
            bfs(s1[i]);
            if(flag==0) break; 
        }
     }
     if(flag==0) puts("NO");
     else puts("YES");
    return 0;
}