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 = 400 + 5;
char s[N], t[N];
int dp[N][N];
vector<char> a, b;
bool DP(int n, int m, int t) {
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= m; j++) dp[i][j] = -1;
  for (int i = 0; i <= n; i++) dp[i][0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= i && j < a.size(); j++) {
      if (dp[i - 1][j] != -1) {
        dp[i][j] = max(dp[i][j], dp[i - 1][j] + (s[i] == b[dp[i - 1][j] + 1]));
      }
      if (s[i] == a[j] && dp[i - 1][j - 1] != -1) {
        dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
      }
    }
  }
  return dp[n][a.size() - 1] >= ((int)b.size() - 1);
}
int main() {
  int T;
  scanf("%d", &T);
  for (; T--;) {
    scanf("%s%s", s + 1, t + 1);
    int n = strlen(s + 1), m = strlen(t + 1);
    int cur = 1;
    for (int i = 1; i <= n; i++) {
      if (s[i] == t[cur]) cur++;
    }
    if (cur > m) {
      puts("YES");
      continue;
    }
    bool flag = 0;
    for (int i = 2; i <= m; i++) {
      a.clear();
      b.clear();
      a.push_back(0);
      b.push_back(0);
      for (int j = 1; j < i; j++) a.push_back(t[j]);
      for (int j = i; j <= m; j++) b.push_back(t[j]);
      if (DP(n, m, i)) {
        flag = 1;
        break;
      }
    }
    puts(flag ? "YES" : "NO");
  }
  return 0;
}
