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 ld = long double;
using namespace std;
const double eps = (1e-11);
const long long INF = 2e17 + 9;
const int logn = 21;
const double pi = acos(-1.0);
int pp = INT_MAX;
long long dp[405][405];
char s[2001];
char t[2001];
long long LCS(char *a1, char *a2, char *a3, long long n, long long m,
              long long l) {
  for (long long i = 0; i <= n; i++) {
    dp[i][0] = 0;
    for (long long j = 1; j <= m; j++) {
      dp[i][j] = -1;
    }
  }
  for (long long i = 1; i <= n; i++) {
    dp[i][0] = dp[i - 1][0];
    if (a1[i - 1] == a3[dp[i - 1][0]]) dp[i][0]++;
    for (long long j = 1; j <= m; j++) {
      long long up = 0;
      if (dp[i - 1][j] != -1) {
        if (a1[i - 1] == a3[dp[i - 1][j]]) {
          up = 1;
        }
        dp[i][j] = dp[i - 1][j] + up;
      }
      if (a1[i - 1] == a2[j - 1]) {
        if (dp[i - 1][j - 1] == -1) {
          dp[i][j] = -1;
        } else if (dp[i - 1][j] != -1)
          dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
        else
          dp[i][j] = dp[i - 1][j - 1];
      }
    }
  }
  if (dp[n][m] == l) return true;
  return false;
}
signed main() {
  long long tt;
  scanf("%lld", &tt);
  while (tt--) {
    scanf("%s", s);
    scanf("%s", t);
    long long n = strlen(s);
    long long m = strlen(t);
    for (long long i = 1; i < m; i++) {
      if (LCS(s, t, t + i, n, i, m - i)) {
        printf("YES\n");
        goto L;
      }
    }
    for (long long i = 0; i <= n; i++) {
      dp[i][0] = 0;
      for (long long j = 1; j <= m; j++) {
        dp[i][j] = 0;
      }
    }
    for (long long i = 1; i <= n; i++) {
      for (long long j = 1; j <= m; j++) {
        dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
        if (s[i - 1] == t[j - 1])
          dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] + 1);
      }
    }
    if (dp[n][m] == m) {
      printf("YES\n");
      goto L;
    }
    printf("NO\n");
  L:;
  }
  return 0;
}
