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;
bool have[405][405][405] = {0};
short int DP[405][405][405];
char a[405];
char b[405];
short N;
short int F(int here, int x, int y) {
  if (a[here] == '\0') {
    if (b[y] == '\0') return x;
    return -1;
  }
  if (have[here][x][y]) return DP[here][x][y];
  have[here][x][y] = 1;
  DP[here][x][y] = F(here + 1, x, y);
  if (a[here] == b[y])
    DP[here][x][y] = max(DP[here][x][y], F(here + 1, x, y + 1));
  if (a[here] == b[x])
    DP[here][x][y] = max(DP[here][x][y], F(here + 1, x + 1, y));
  return DP[here][x][y];
}
int main() {
  int T, ok, i, j, k;
  scanf("%d", &T);
  while (T--) {
    scanf("%s %s", a, b);
    ok = 0;
    N = strlen(a);
    for (i = 0; i <= N; i++)
      for (j = 0; j <= N; j++)
        for (k = 0; k <= N; k++) have[i][j][k] = 0;
    for (i = 0; i < N; i++) {
      if (F(0, 0, i) == i) ok = 1;
    }
    if (ok)
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}
