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 maxn = 2e5 + 22;
int T, l1, l2, dp[500][500];
char s[maxn], t[maxn], a[maxn], b[maxn];
int nextc[maxn][30];
void init_xulie() {
  for (int j = 0; j <= 25; j++) nextc[l1][j] = mod;
  for (int i = l1; i >= 1; i--) {
    for (int j = 0; j < 26; j++) {
      nextc[i - 1][j] = nextc[i][j];
    }
    nextc[i - 1][s[i] - 'a'] = i;
  }
}
int gkd(int x) {
  int n = x, m = l2 - x;
  for (int i = 1; i <= n; i++) a[i] = t[i];
  for (int i = 1; i <= m; i++) b[i] = t[n + i];
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= m; j++) dp[i][j] = mod;
  dp[0][0] = 0;
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= m; j++) {
      if (dp[i][j] == mod) continue;
      dp[i + 1][j] = min(dp[i + 1][j], nextc[dp[i][j]][a[i + 1] - 'a']);
      dp[i][j + 1] = min(dp[i][j + 1], nextc[dp[i][j]][b[j + 1] - 'a']);
    }
  }
  return dp[n][m];
}
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%s%s", s + 1, t + 1);
    int ans = mod;
    l1 = strlen(s + 1);
    l2 = strlen(t + 1);
    init_xulie();
    for (int i = 0; i <= l2; i++) ans = min(ans, gkd(i));
    ;
    if (ans <= l1)
      puts("YES");
    else
      puts("NO");
  }
  return 0;
}
