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 read() {
  int xx = 0, ff = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') ff = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    xx = xx * 10 + ch - '0';
    ch = getchar();
  }
  return xx * ff;
}
long long READ() {
  long long xx = 0, ff = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') ff = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    xx = xx * 10 + ch - '0';
    ch = getchar();
  }
  return xx * ff;
}
char one() {
  char ch = getchar();
  while (ch == ' ' || ch == '\n') ch = getchar();
  return ch;
}
const int maxn = 410;
char s[maxn], t[maxn];
int n, m, pos[26], nxt[maxn][26];
int f[maxn][maxn];
int main() {
  for (int _ = read(); _; _--) {
    scanf("%s", s + 1);
    scanf("%s", t + 1);
    n = strlen(s + 1);
    m = strlen(t + 1);
    memset(pos, 10, sizeof(pos));
    for (int i = n; i >= 1; i--) {
      pos[s[i] - 'a'] = i;
      for (int j = 0; j < 26; j++) nxt[i][j] = pos[j];
    }
    bool yes = 0;
    for (int p = 1; p <= m; p++) {
      memset(f, 10, sizeof(f));
      f[0][p - 1] = 0;
      for (int i = 0; i < p; i++)
        for (int j = p - 1; j <= m; j++) {
          int tmp = f[i][j];
          if (f[i][j] > n) break;
          if (i != p - 1 && tmp < n)
            f[i + 1][j] = min(f[i + 1][j], nxt[tmp + 1][t[i + 1] - 'a']);
          if (j != m && tmp < n)
            f[i][j + 1] = min(f[i][j + 1], nxt[tmp + 1][t[j + 1] - 'a']);
        }
      if (f[p - 1][m] <= n) {
        yes = 1;
        break;
      }
    }
    if (yes)
      puts("YES");
    else
      puts("NO");
  }
  return 0;
}
