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;
int T, n, m, ps[405][26], dp[405][405];
bool fl;
char a[405], b[405];
int min(int x, int y) { return x < y ? x : y; }
bool chk(int x) {
  memset(dp, 0, sizeof(dp));
  for (int i = 0; i <= x; ++i)
    for (int j = 0; j <= m - x; ++j) {
      dp[i][j] = 0x3f3f3f3f;
      if (!i && !j) dp[i][j] = 0;
      if (i && dp[i - 1][j] < 0x3f3f3f3f && ps[dp[i - 1][j]][b[i] - 'a'])
        dp[i][j] = ps[dp[i - 1][j]][b[i] - 'a'];
      if (j && dp[i][j - 1] < 0x3f3f3f3f && ps[dp[i][j - 1]][b[j + x] - 'a'])
        dp[i][j] = min(dp[i][j], ps[dp[i][j - 1]][b[j + x] - 'a']);
    }
  return dp[x][m - x] < 0x3f3f3f3f;
}
int main() {
  scanf("%d", &T);
  while (T--) {
    fl = 0;
    memset(ps, 0, sizeof(ps));
    scanf("%s %s", a + 1, b + 1);
    n = strlen(a + 1);
    m = strlen(b + 1);
    for (int i = n - 1; i >= 0; --i) {
      ps[i][a[i + 1] - 'a'] = i + 1;
      for (int j = 0; j < 26; ++j)
        if (j != a[i + 1] - 'a') ps[i][j] = ps[i + 1][j];
    }
    for (int i = 1; i < m; ++i)
      if (chk(i)) {
        fl = 1;
        puts("YES");
        break;
      }
    if (fl) continue;
    if (m > 1) {
      puts("NO");
      continue;
    }
    for (int i = 1; i <= n; ++i)
      if (a[i] == b[1]) {
        fl = 1;
        puts("YES");
        break;
      }
    if (!fl) puts("NO");
  }
}
