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;
char a[444], b[444];
vector<vector<vector<bool>>> vis;
int p, m, n;
int dp(int p1, int p2, int x) {
  if (p1 == p && p2 == m) {
    return 1;
  }
  if (x == n) {
    return 0;
  }
  if (vis[x][p1][p2]) return 0;
  int res = 0;
  if (p1 < p && a[x] == b[p1]) {
    res = dp(p1 + 1, p2, x + 1);
  }
  if (res) return 1;
  if (p2 < m && a[x] == b[p2]) {
    res = dp(p1, p2 + 1, x + 1);
  }
  if (res) return 1;
  res = dp(p1, p2, x + 1);
  if (res) return 1;
  vis[x][p1][p2] = 1;
  return 0;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long ans = 0;
    scanf("%s%s", a, b);
    n = strlen(a);
    m = strlen(b);
    bool tmp = 0;
    vis = vector<vector<vector<bool>>>(
        n + 1, vector<vector<bool>>(m + 1, vector<bool>(m + 1, 0)));
    for (p = 0; p <= m; ++p) {
      tmp = dp(0, p, 0);
      if (tmp) {
        printf("YES\n");
        break;
      }
    }
    if (!tmp) {
      printf("NO\n");
    }
  }
  return 0;
}
