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 N = 400;
const int inf = 1e3;
int dp[N + 10][N + 10];
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    char s[N + 10], t[N + 10];
    scanf("%s%s", s, t);
    int szs = strlen(s);
    int szt = strlen(t);
    int nxt[N + 10][30];
    for (char j = 'a'; j <= 'z'; j++) {
      nxt[szs][j - 'a'] = szs + 1;
    }
    for (int i = szs - 1; i >= 0; i--) {
      for (char j = 'a'; j <= 'z'; j++) {
        if (s[i] == j) {
          nxt[i][j - 'a'] = i + 1;
        } else {
          nxt[i][j - 'a'] = nxt[i + 1][j - 'a'];
        }
      }
    }
    bool flag = false;
    for (int i = 0; i < szt; i++) {
      string s1 = "", s2 = "";
      for (int j = 0; j <= i; j++) {
        s1 += t[j];
      }
      for (int j = i + 1; j < szt; j++) {
        s2 += t[j];
      }
      int n = s1.size(), m = s2.size();
      for (int j = 0; j <= n + 1; j++) {
        for (int k = 0; k <= m + 1; k++) {
          dp[j][k] = inf;
        }
      }
      dp[0][0] = 0;
      for (int j = 0; j <= n; j++) {
        for (int k = 0; k <= m; k++) {
          int cur = dp[j][k];
          if (cur >= szs) {
            continue;
          }
          if (j != n) {
            int nx = nxt[cur][s1[j] - 'a'];
            dp[j + 1][k] = min(dp[j + 1][k], nx);
          }
          if (k != m) {
            int nx = nxt[cur][s2[k] - 'a'];
            dp[j][k + 1] = min(dp[j][k + 1], nx);
          }
        }
      }
      if (dp[n][m] <= szs) {
        flag = true;
      }
    }
    if (flag == true) {
      printf("YES\n");
    } else {
      printf("NO\n");
    }
  }
}
