Key

/*
    Task    : _example
    Author  : Phumipat C. [MAGCARI]
    Language: C++
    Created : 31 December 2022 [11:49]
*/
#include<bits/stdc++.h>
using namespace std;
char a[1010],b[1010],key[2010];
int dp[1010][1010]; // consider a[1..i] and b[1..j] can make key[1..i+j]
int main(){
    cin.tie(0)->sync_with_stdio(0);
    cin.exceptions(cin.failbit);
    cin >> a+1 >> b+1;
    int n = strlen(a+1),m = strlen(b+1);
    int q;
    cin >> q;
    while(q--){
        cin >> key+1;
        dp[0][0] = 1;
        for(int i=1;i<=n;i++)
            if(a[i] == key[i] && dp[i-1][0])
                dp[i][0] = 1;
        for(int i=1;i<=m;i++)
            if(b[i] == key[i] && dp[0][i-1])
                dp[0][i] = 1;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
                if((a[i] == key[i+j] && dp[i-1][j]) || (b[j] == key[i+j] && dp[i][j-1]))
                    dp[i][j] = 1;
        if(dp[n][m])    cout << "Yes\n";
        else            cout << "No\n";
        memset(dp,0,sizeof dp);
    }
    return 0;
}