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 T;
string s, t, a, b;
int nxt[405][501];
int dp[405][405];
void reset() {
  for (int j = 0; j <= 400; j++)
    for (int l = 0; l <= 400; l++) dp[j][l] = -5;
}
bool isok() {
  reset();
  dp[0][0] = -1;
  for (int lga = 0; lga <= a.size(); lga++)
    for (int lgb = 0; lgb <= b.size(); lgb++)
      if (dp[lga][lgb] == -5) {
        if (lga > 0 && dp[lga - 1][lgb] != -5) {
          int pos = dp[lga - 1][lgb] + 1;
          char chr = a[lga - 1];
          if (nxt[pos][int(chr)] != -5) {
            if (dp[lga][lgb] == -5)
              dp[lga][lgb] = nxt[pos][chr];
            else
              dp[lga][lgb] = min(dp[lga][lgb], nxt[pos][chr]);
          }
        }
        if (lgb > 0 && dp[lga][lgb - 1] != -5) {
          int pos = dp[lga][lgb - 1] + 1;
          char chr = b[lgb - 1];
          if (nxt[pos][chr] != -5) {
            if (dp[lga][lgb] == -5)
              dp[lga][lgb] = nxt[pos][chr];
            else
              dp[lga][lgb] = min(dp[lga][lgb], nxt[pos][chr]);
          }
        }
      }
  if (dp[a.size()][b.size()] != -5)
    return 1;
  else
    return 0;
}
void solve() {
  cin >> s >> t;
  for (int i = 0; i <= 400; i++)
    for (int j = 'a'; j <= 'z'; j++) nxt[i][j] = -5;
  for (int i = 0; i < s.size(); i++)
    for (int j = i; j < s.size(); j++)
      if (nxt[i][s[j]] == -5) nxt[i][s[j]] = j;
  bool ok = 0;
  int as = 0;
  while (as <= t.size()) {
    a = t.substr(0, as);
    b = t.substr(as);
    if (isok()) {
      ok = 1;
      break;
    }
    as++;
  }
  if (ok)
    cout << "YES" << '\n';
  else
    cout << "NO" << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> T;
  for (int q = 1; q <= T; q++) solve();
  return 0;
}
