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>
const long long oo = (long long)1e18 + 1;
using namespace std;
int nt, n, m;
string S, T;
int nxt[555][30];
int dp[555][555];
void init() {
  for (int i = 0; i < 26; i++) {
    nxt[n + 1][i] = n + 1;
    nxt[n + 2][i] = n + 1;
  }
  for (int i = n; i >= 1; i--) {
    int c = S[i - 1] - 'a';
    for (int j = 0; j < 26; j++) {
      if (j == c)
        nxt[i][j] = i;
      else
        nxt[i][j] = nxt[i + 1][j];
    }
  }
}
bool go_dp(int mid) {
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= n; j++) dp[i][j] = n + 1;
  dp[0][0] = 0;
  for (int i = 1; i <= mid; i++) {
    int ch = T[i - 1] - 'a';
    dp[i][0] = nxt[dp[i - 1][0] + 1][ch];
  }
  for (int i = mid + 1; i <= m; i++) {
    int ch = T[i - 1] - 'a';
    int j = i - mid;
    dp[0][j] = nxt[dp[0][j - 1] + 1][ch];
  }
  for (int i = 1; i <= mid; i++)
    for (int k = mid + 1; k <= m; k++) {
      int j = k - mid;
      int ch_fi = T[i - 1] - 'a';
      int ch_se = T[k - 1] - 'a';
      dp[i][j] =
          min(nxt[dp[i - 1][j] + 1][ch_fi], nxt[dp[i][j - 1] + 1][ch_se]);
    }
  return (dp[mid][m - mid] <= n);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> nt;
  while (nt--) {
    cin >> S;
    cin >> T;
    n = S.size();
    m = T.size();
    if (m > n) {
      cout << "NO\n";
      continue;
    }
    init();
    bool could = false;
    for (int k = 0; k <= m; k++) {
      if (go_dp(k)) {
        could = true;
        break;
      }
    }
    if (could)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}
