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 mod = 1000000007;
;
const int maxn = 410;
unsigned long long powmod(unsigned long long n, unsigned long long k) {
  unsigned long long ret = 1;
  while (k) {
    if (k & 1) ret = ret * n % mod;
    n = n * n % mod;
    k >>= 1;
  }
  return ret;
}
int nxt[maxn][26];
int dp[maxn][maxn];
void upd(int& n, int k) {
  if (n < 0 || n > k) n = k;
}
bool gao(const string& s, const string& t1, const string& t2) {
  memset(dp, -1, sizeof(dp));
  for (int i = 0; i <= t1.size(); i++)
    for (int j = 0; j <= t2.size(); j++)
      if (dp[i][j] >= 0 || i + j == 0) {
        int p = dp[i][j] + 1;
        if (i < t1.size()) {
          int n = nxt[p][t1[i] - 'a'];
          if (n >= 0) upd(dp[i + 1][j], n);
        }
        if (j < t2.size()) {
          int n = nxt[p][t2[j] - 'a'];
          if (n >= 0) upd(dp[i][j + 1], n);
        }
      }
  return dp[t1.size()][t2.size()] >= 0;
}
char str[maxn];
char tt[maxn];
int main() {
  int t;
  cin >> t;
  while (t--) {
    cin >> str >> tt;
    memset(nxt, -1, sizeof(nxt));
    for (int i = strlen(str) - 1; i >= 0; i--) {
      memcpy(nxt[i], nxt[i + 1], sizeof(nxt[i]));
      nxt[i][str[i] - 'a'] = i;
    }
    bool yes = false;
    if (strlen(tt) == 1)
      yes = (nxt[0][tt[0] - 'a'] >= 0);
    else {
      for (int i = 0; tt[i + 1]; i++) {
        char tmp = tt[i + 1];
        tt[i + 1] = 0;
        string t1 = tt;
        tt[i + 1] = tmp;
        string t2 = tt + i + 1;
        if (gao(str, t1, t2)) {
          yes = true;
          break;
        }
      }
    }
    if (yes)
      puts("YES");
    else
      puts("NO");
  }
}
