Subsequence （暴力搜索）

Give a string SSS and NNN string TiT_iTi​ , determine whether TiT_iTi​ is a subsequence of SSS.

If ti is subsequence of SSS, print YES,else print NO.

If there is an array {K1,K2,K3,⋯ ,Km}lbrace K_1, K_2, K_3,cdots, K_m brace{K1​,K2​,K3​,⋯,Km​} so that 1≤K1<K2<K3<⋯<Km≤N1 le K_1 < K_2 < K_3 < cdots < K_m le N1≤K1​<K2​<K3​<⋯<Km​≤N and Ski=TiS_{k_i} = T_iSki​​=Ti​, (1≤i≤m)(1 le i le m)(1≤i≤m), then TiT_iTi​ is a subsequence of SSS.

Input

The first line is one string SSS,length(SSS) ≤100000 le 100000≤100000

The second line is one positive integer N,N≤100000N,N le 100000N,N≤100000

Then next nnn lines,every line is a string TiT_iTi​, length(TiT_iTi​) ≤1000le 1000≤1000

Output

Print NNN lines. If the iii-th TiT_iTi​ is subsequence of SSS, print YES, else print NO.

样例输入 

abcdefg
3
abc
adg
cba

样例输出 

YES
YES
NO

#include <cstdio>
#include <iostream>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
#define ll long long
const int inf = 0x3f3f3f3f;
const  ll linf  =1LL<<50;
const int maxn = 1e5+8;
string s, miao;
int n;
int main()
{
    cin>>s;
    int l = s.size();
    cin >> n;
    while(n--)
    {
        cin>>miao;
        int len = miao.size();
        int ga = 0;
        for(int i = 0; i<l; i++)
        {
            if(miao[ga] == s[i])
            {
                ga++;
            }
            if(ga == len)break;
        }
        if(ga == len)printf("YES
");
        else printf("NO
");
    }
    return 0;
}