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 = 410;
int n, m;
int nxt[N][26];
char s[N], t[N];
void init() {
  for (int j = 0; j <= 25; j++) {
    nxt[n][j] = n + 1;
    nxt[n + 1][j] = n + 1;
  }
  for (int i = n - 1; i >= 0; i--) {
    for (int j = 0; j <= 25; j++) nxt[i][j] = nxt[i + 1][j];
    nxt[i][s[i + 1] - 'a'] = i + 1;
  }
}
char t1[N], t2[N];
int F[N][N];
bool Calc(int n1, int n2) {
  for (int i = 1; i <= n1; i++) t1[i] = t[i];
  for (int i = 1; i <= n2; i++) t2[i] = t[i + n1];
  for (int i = 0; i <= n1; i++)
    for (int j = 0; j <= n2; j++) F[i][j] = n + 1;
  F[0][0] = 0;
  for (int i = 0; i <= n1; i++)
    for (int j = 0; j <= n2; j++) {
      if (i > 0) F[i][j] = min(F[i][j], nxt[F[i - 1][j]][t1[i] - 'a']);
      if (j > 0) F[i][j] = min(F[i][j], nxt[F[i][j - 1]][t2[j] - 'a']);
    }
  if (F[n1][n2] <= n) return 1;
  return 0;
}
int main() {
  int tCase;
  cin >> tCase;
  while (tCase--) {
    cin >> (s + 1);
    n = strlen(s + 1);
    cin >> (t + 1);
    m = strlen(t + 1);
    init();
    bool ok = 0;
    for (int n1 = 1; n1 <= m; n1++)
      if (Calc(n1, m - n1)) {
        cout << "YES" << endl;
        ok = 1;
        break;
      }
    if (!ok) cout << "NO" << endl;
  }
}
