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;
int T, n, m, nxt[MAXN][26], dp[MAXN][MAXN];
char s[MAXN], t[MAXN];
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%s %s", s, t);
    n = strlen(s);
    m = strlen(t);
    for (int i = n; i >= 0; i--)
      for (int j = 0; j < 26; j++) {
        if (i == n)
          nxt[i][j] = -1;
        else {
          if (s[i] - 'a' == j)
            nxt[i][j] = i;
          else
            nxt[i][j] = nxt[i + 1][j];
        }
      }
    bool ok = false;
    for (int sep = 0; sep < m; sep++) {
      for (int i = 0; i <= sep; i++)
        for (int j = 0; j <= m - sep; j++) {
          if (i == 0 && j == 0)
            dp[i][j] = 0;
          else {
            dp[i][j] = -1;
            if (i > 0 && dp[i - 1][j] != -1) {
              if (nxt[dp[i - 1][j]][t[i - 1] - 'a'] != -1) {
                dp[i][j] = nxt[dp[i - 1][j]][t[i - 1] - 'a'] + 1;
              }
            }
            if (j > 0 && dp[i][j - 1] != -1) {
              if (nxt[dp[i][j - 1]][t[sep + j - 1] - 'a'] != -1) {
                if (dp[i][j] == -1 ||
                    dp[i][j] > nxt[dp[i][j - 1]][t[sep + j - 1] - 'a'] + 1) {
                  dp[i][j] = nxt[dp[i][j - 1]][t[sep + j - 1] - 'a'] + 1;
                }
              }
            }
          }
        }
      if (dp[sep][m - sep] != -1) {
        ok = true;
        break;
      }
    }
    if (ok)
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}
