You are given a string s. You can build new string p from s using the following operation no more than two times: 

  1. choose any subsequence s_{i_1}, s_{i_2}, ..., s_{i_k} where 1 ≤ i_1 < i_2 < ... < i_k ≤ |s|; 
  2. erase the chosen subsequence from s (s can become empty); 
  3. concatenate chosen subsequence to the right of the string p (in other words, p = p + s_{i_1}s_{i_2}... s_{i_k}). 



Of course, initially the string p is empty. 

For example, let s = ababcd. At first, let's choose subsequence s_1 s_4 s_5 = abc — we will get s = bad and p = abc. At second, let's choose s_1 s_2 = ba — we will get s = d and p = abcba. So we can build abcba from ababcd.

Can you build a given string t using the algorithm above?

Input

The first line contains the single integer T (1 ≤ T ≤ 100) — the number of test cases.

Next 2T lines contain test cases — two per test case. The first line contains string s consisting of lowercase Latin letters (1 ≤ |s| ≤ 400) — the initial string.

The second line contains string t consisting of lowercase Latin letters (1 ≤ |t| ≤ |s|) — the string you'd like to build.

It's guaranteed that the total length of strings s doesn't exceed 400.

Output

Print T answers — one per test case. Print YES (case insensitive) if it's possible to build t and NO (case insensitive) otherwise.

Example

Input


4
ababcd
abcba
a
b
defi
fed
xyz
x


Output


YES
NO
NO
YES
#include <bits/stdc++.h>
using namespace std;
char s[407], a[407];
int dp[407][407];
int ls, la, ta, oa, ob;
char geta(int i) { return a[i]; }
char getb(int i) { return a[i + ta]; }
int nxt[407][26], nt[26];
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%s%s", s + 1, a + 1);
    ls = strlen(s + 1);
    la = strlen(a + 1);
    for (int j = 0; j < 26; j++) {
      nxt[405][j] = nt[j] = 405;
    }
    for (int i = ls; i >= 1; i--) {
      for (int j = 0; j < 26; j++) {
        nxt[i][j] = nt[j];
      }
      nt[s[i] - 'a'] = i;
    }
    for (int j = 0; j < 26; j++) {
      nxt[0][j] = nt[j];
    }
    int flag = 0;
    for (ta = 0; ta <= la; ta++) {
      oa = ta;
      ob = la - ta;
      for (int i = 1; i <= oa; i++) {
        dp[i][0] = nxt[dp[i - 1][0]][geta(i) - 'a'];
      }
      for (int j = 1; j <= ob; j++) {
        dp[0][j] = nxt[dp[0][j - 1]][getb(j) - 'a'];
      }
      for (int i = 1; i <= oa; i++) {
        for (int j = 1; j <= ob; j++) {
          dp[i][j] = min(nxt[dp[i - 1][j]][geta(i) - 'a'],
                         nxt[dp[i][j - 1]][getb(j) - 'a']);
        }
      }
      if (dp[oa][ob] != 405) {
        flag = 1;
        break;
      }
    }
    if (flag) {
      puts("YES");
    } else {
      puts("NO");
    }
  }
}
