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;
const int maxn = 405;
const int inf = 0x3f3f3f3f;
char s[maxn], t[maxn];
char t1[maxn], t2[maxn];
int f[maxn][maxn];
int ch[maxn][30];
bool check() {
  int t1_len = (int)strlen(t1 + 1);
  int t2_len = (int)strlen(t2 + 1);
  for (int i = 0; i <= t1_len; i++) {
    for (int j = 0; j <= t2_len; j++) {
      if (i == 0 && j == 0) {
        f[i][j] = 0;
      } else {
        f[i][j] = inf;
      }
      if (i > 0 && f[i - 1][j] < inf) {
        int pos = f[i - 1][j];
        pos = ch[pos][t1[i] - 'a'];
        f[i][j] = min(f[i][j], pos);
      }
      if (j > 0 && f[i][j - 1] < inf) {
        int pos = f[i][j - 1];
        pos = ch[pos][t2[j] - 'a'];
        f[i][j] = min(f[i][j], pos);
      }
    }
  }
  return f[t1_len][t2_len] <= strlen(s + 1);
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%s%s", s + 1, t + 1);
    int len = (int)strlen(s + 1);
    int pos[30];
    for (int i = 0; i < 26; i++) pos[i] = inf;
    for (int i = len; i >= 0; i--) {
      for (int j = 0; j < 26; j++) {
        ch[i][j] = pos[j];
      }
      int x = s[i] - 'a';
      if (i > 0) pos[x] = i;
    }
    len = (int)strlen(t + 1);
    bool ok = false;
    for (int i = 1; i <= len; i++) {
      for (int j = 1; j < i; j++) {
        t1[j] = t[j];
      }
      t1[i] = '\0';
      for (int j = i; j <= len; j++) {
        t2[j - i + 1] = t[j];
      }
      t2[len - i + 2] = '\0';
      if (check()) {
        ok = true;
        break;
      }
    }
    if (ok) {
      printf("YES\n");
    } else {
      printf("NO\n");
    }
  }
  return 0;
}
