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 mod = 1e9 + 7;
const int N = 404;
int power(int b, int exp) {
  int res = 1;
  while (exp > 0) {
    if (exp % 2) {
      res = 1LL * res * b % mod;
    }
    b = 1LL * b * b % mod;
    exp /= 2;
  }
  return res;
}
int n, m, last;
char s[N], t[N];
int dp[N][N][2];
int fun(int idx, int nxt, int to_pick) {
  if (idx == n + 1) {
    if (nxt > last) {
      return 0;
    }
    return -1000;
  }
  int &ret = dp[idx][nxt][to_pick];
  if (ret != -1) {
    return ret;
  }
  ret = fun(idx + 1, nxt, to_pick);
  if (nxt <= last && s[idx] == t[nxt]) {
    ret = max(ret, fun(idx + 1, nxt + 1, to_pick));
  }
  if (to_pick && s[idx] == t[m]) {
    ret = max(ret, 1 + fun(idx + 1, nxt, 0));
  }
  if (to_pick) {
    int len = fun(idx + 1, nxt, to_pick);
    if (len >= 0 && m - len > 0 && s[idx] == t[m - len]) {
      ret = max(ret, 1 + len);
    }
  }
  return ret;
}
int main() {
  int test;
  scanf("%d", &test);
  while (test--) {
    int yes = 0;
    scanf("%s", s + 1);
    n = strlen(s + 1);
    scanf("%s", t + 1);
    m = strlen(t + 1);
    for (int i = 1; i <= m; i++) {
      last = i;
      for (int j = 0; j <= max(n, m); j++) {
        for (int k = 0; k <= (n, m); k++) {
          for (int l = 0; l < 2; l++) {
            dp[j][k][l] = -1;
          }
        }
      }
      if (fun(1, 1, 1) >= m - i) {
        yes = 1;
        break;
      }
    }
    if (yes) {
      printf("YES\n");
    } else {
      printf("NO\n");
    }
  }
  return 0;
}
