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 maxn = 1e5 + 7;
const int mod = 1e9 + 7;
const long long INF = 1e18;
int dp[405][405][405];
char a[405], b[405];
int sta, stb;
int check(int x, int y, int z) {
  if (z >= sta) {
    if (y >= stb) return x;
    return -10;
  }
  if (dp[x][y][z] != -1) return dp[x][y][z];
  dp[x][y][z] = check(x, y, z + 1);
  if (x < stb && a[z] == b[x])
    dp[x][y][z] = max(dp[x][y][z], check(x + 1, y, z + 1));
  if (y < stb && a[z] == b[y])
    dp[x][y][z] = max(dp[x][y][z], check(x, y + 1, z + 1));
  return dp[x][y][z];
}
bool rua() {
  scanf("%s", a);
  sta = strlen(a);
  scanf("%s", b);
  stb = strlen(b);
  for (int i = 0; i <= stb; i++)
    for (int j = 0; j <= stb; j++)
      for (int k = 0; k <= sta; k++) dp[i][j][k] = -1;
  for (int i = 0; i < stb; i++)
    if (check(0, i, 0) >= i) return 1;
  return 0;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    if (rua())
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}
