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 = 405;
char s[N], t[N];
char s1[N], s2[N];
int n, m, l1, l2;
int f[N][N];
bool check(int l) {
  for (int i = 0; i < m; i++)
    if (s[l + i] != t[i]) return false;
  return true;
}
bool dp() {
  memset(f, -1, sizeof(f));
  f[0][0] = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= l1; j++) {
      if (f[i][j] >= 0) {
        int k = f[i][j];
        if (s[i] == s1[j] && j < l1) f[i + 1][j + 1] = max(k, f[i + 1][j + 1]);
        if (s[i] == s2[k] && k < l2) f[i + 1][j] = max(f[i + 1][j], k + 1);
        f[i + 1][j] = max(f[i + 1][j], f[i][j]);
      }
    }
  }
  if (f[n][l1] == l2) return true;
  return false;
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    cin >> s >> t;
    n = strlen(s);
    m = strlen(t);
    bool flag = false;
    for (int i = 0; i < n; i++) {
      if (s[i] == t[0]) {
        if (check(i)) {
          flag = true;
          break;
        }
      }
    }
    if (flag) {
      printf("YES\n");
      continue;
    }
    flag = false;
    for (int mid = 0; mid < m - 1; mid++) {
      for (int i = 0; i < N; i++) s1[i] = s2[i] = ' ';
      l1 = l2 = 0;
      for (int i = 0; i <= mid; i++) s1[l1++] = t[i];
      for (int i = mid + 1; i < m; i++) s2[l2++] = t[i];
      if (dp()) {
        flag = true;
        break;
      }
    }
    if (flag) {
      printf("YES\n");
      continue;
    } else
      printf("NO\n");
  }
  return 0;
}
