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;
char T[500];
char s[500];
int dp[405][405];
int tms[405][405];
int seq = 0;
int len, L;
bool tryit(int x) {
  seq++;
  memset(dp, 0, sizeof(dp));
  memset(tms, 0, sizeof(tms));
  dp[0][0] = 0;
  tms[0][0] = seq;
  for (int i = 1; i <= len; ++i)
    for (int j = 0; j <= x; ++j) {
      if (tms[i - 1][j] != seq) break;
      dp[i][j] = max(dp[i][j], dp[i - 1][j]);
      ;
      tms[i][j] = tms[i - 1][j];
      if (T[i] == s[j + 1]) {
        tms[i][j + 1] = tms[i - 1][j];
        dp[i][j + 1] = max(dp[i][j + 1], dp[i - 1][j]);
        ;
      }
      if (dp[i][j] + x + 1 <= L && T[i] == s[x + dp[i][j] + 1]) {
        tms[i][j] = tms[i - 1][j];
        dp[i][j] = max(dp[i][j], dp[i - 1][j] + 1);
        ;
      }
    }
  if (tms[len][x] == seq && dp[len][x] + x == L)
    return 1;
  else
    return 0;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%s", T + 1);
    len = strlen(T + 1);
    scanf("%s", s + 1);
    L = strlen(s + 1);
    bool flag = false;
    for (int i = 1; i <= len; ++i) {
      if (tryit(i)) {
        flag = 1;
        break;
      }
    }
    if (flag)
      puts("YES");
    else
      puts("NO");
  }
  return 0;
}
