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;
int n, a[3001], b[3001][26], t;
int dp[3001][3001];
char a1[3010], a2[3010];
int main() {
  cin >> t;
  for (int yui = 1; yui <= t; yui++) {
    int biao = 1;
    cin >> (a1 + 1) >> (a2 + 1);
    int len1, len2;
    len1 = strlen(a1 + 1);
    len2 = strlen(a2 + 1);
    for (int opi = 406; opi >= 0; opi--) {
      for (int yu = 0; yu <= 25; yu++) {
        b[opi][yu] = 405;
      }
    }
    for (int opi = len1; opi >= 0; opi--) {
      int pos;
      pos = a1[opi] - 'a';
      for (int yu = 0; yu <= 25; yu++) {
        if (b[opi][yu] == 405) b[opi][yu] = b[opi + 1][yu];
        if (yu == pos) {
          b[opi - 1][yu] = opi;
        }
      }
    }
    for (int opi = 1; opi <= len2; opi++) {
      for (int l1 = 0; l1 <= opi; l1++) {
        for (int r1 = 0; r1 <= len2 - opi; r1++) {
          if (l1 == 0 && r1 == 0) {
            dp[l1][r1] = 0;
            continue;
          }
          if (l1 == 0) {
            dp[l1][r1] = b[dp[l1][r1 - 1]][a2[r1 + opi] - 'a'];
            continue;
          }
          if (r1 == 0) {
            dp[l1][r1] = b[dp[l1 - 1][r1]][a2[l1] - 'a'];
            continue;
          }
          dp[l1][r1] = min(b[dp[l1 - 1][r1]][a2[l1] - 'a'],
                           b[dp[l1][r1 - 1]][a2[opi + r1] - 'a']);
        }
      }
      if (dp[opi][len2 - opi] != 405) {
        biao = 0;
        cout << "YES" << endl;
        break;
      }
    }
    if (biao == 1) {
      cout << "NO" << endl;
    }
  }
}
