Educational Codeforces Round 68 (Rated for Div. 2)-C-From S To T

You are given three strings ss, tt and pp consisting of lowercase Latin letters. You may perform any number (possibly, zero) operations on these strings.

During each operation you choose any character from pp, erase it from pp and insert it into string ss (you may insert this character anywhere you want: in the beginning of ss, in the end or between any two consecutive characters).

For example, if pp is aba, and ss is de, then the following outcomes are possible (the character we erase from pp and insert into ss is highlighted):

• aba →→ ba, de →→ ade;
• aba →→ ba, de →→ dae;
• aba →→ ba, de →→ dea;
• aba →→ aa, de →→ bde;
• aba →→ aa, de →→ dbe;
• aba →→ aa, de →→ deb;
• aba →→ ab, de →→ ade;
• aba →→ ab, de →→ dae;
• aba →→ ab, de →→ dea;

Your goal is to perform several (maybe zero) operations so that ss becomes equal to tt. Please determine whether it is possible.

Note that you have to answer qq independent queries.

Input

The first line contains one integer qq (1≤q≤1001≤q≤100) — the number of queries. Each query is represented by three consecutive lines.

The first line of each query contains the string ss (1≤|s|≤1001≤|s|≤100) consisting of lowercase Latin letters.

The second line of each query contains the string tt (1≤|t|≤1001≤|t|≤100) consisting of lowercase Latin letters.

The third line of each query contains the string pp (1≤|p|≤1001≤|p|≤100) consisting of lowercase Latin letters.

Output

For each query print YES if it is possible to make ss equal to tt, and NO otherwise.

You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).

Example

input

Copy

4
ab
acxb
cax
a
aaaa
aaabbcc
a
aaaa
aabbcc
ab
baaa
aaaaa

output

Copy

YES
YES
NO
NO

Note

In the first test case there is the following sequence of operation:

1. s=s= ab, t=t= acxb, p=p= cax;
2. s=s= acb, t=t= acxb, p=p= ax;
3. s=s= acxb, t=t= acxb, p=p= a.

In the second test case there is the following sequence of operation:

1. s=s= a, t=t= aaaa, p=p= aaabbcc;
2. s=s= aa, t=t= aaaa, p=p= aabbcc;
3. s=s= aaa, t=t= aaaa, p=p= abbcc;
4. s=s= aaaa, t=t= aaaa, p=p= bbcc.

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<vector>
#include<map>
#include<cmath>
const int maxn=1e5+;
typedef long long ll;
using namespace std;
string s,t,p;
int vis[maxn];
 
int dp[][];
string str;
int main()
{

   int T;
   cin>>T;
   while(T--)
   {
       cin>>s>>t>>p;
       str=s+p;
       memset(vis,,sizeof(vis));
       int len1=t.length();
       int len2=str.length();
       int len3=s.length();
       int ans=;
       memset(dp,,sizeof(dp));
       for(int i=;i<=len1;i++)
       {
           for(int j=;j<=len3;j++)
           {
               if(t[i-]==s[j-])
               {
                   dp[i][j]=max(dp[i][j],dp[i-][j-]+);
            }
            else
            {
                dp[i][j]=max(dp[i-][j],dp[i][j-]);
            }
        }
    }
    int flag=;
    if(dp[len1][len3]==len3)
    {
        flag=;
    }
    if(flag==)
    {
        puts("NO");
        continue;
    }
    else
    {
        for(int i=;i<len1;i++)
        {
            for(int j=;j<len2;j++)
            {
                if(str[j]==t[i]&&vis[j]==)
                {
                    vis[j]=;
                    ans++;
                    break;
                }
            }
        }
        if(ans==len1)
    {
        puts("YES");
    }
    else
    {
        puts("NO");
    }
    }

   }
   return ;
}