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 N = 405;
const int inf = 1000 * 1000 * 1000;
const int mod = 998244353;
mt19937 myrand(chrono::steady_clock::now().time_since_epoch().count());
int T;
string s, t;
int mx1[N], mx2[N];
int m1[N], m2[N];
bool check() {
  int it = 0;
  for (int i = 0; i < (int)s.size(); i++) {
    if (s[i] == t[it]) {
      ++it;
    }
    if (it == (int)t.size()) return true;
  }
  return false;
}
int main() {
  cin >> T;
  while (T--) {
    cin >> s;
    cin >> t;
    if (check()) {
      printf("YES\n");
      continue;
    }
    bool flag = false;
    for (int i = 0; i < (int)t.size() - 1; i++) {
      for (int j = 0; j <= (int)t.size(); j++) {
        mx1[j] = mx2[j] = -1;
        m1[j] = m2[j] = -1;
      }
      mx1[0] = mx2[0] = 0;
      for (int j = 0; j < (int)s.size(); j++) {
        for (int h = 0; h <= min(j, i); h++) {
          if (s[j] == t[h]) {
            m1[h + 1] = max(m1[h + 1], mx1[h]);
            m2[mx1[h]] = max(m2[mx1[h]], h + 1);
          }
        }
        for (int h = i + 1; h <= min(i + 1 + j, (int)s.size() - 1); h++) {
          if (s[j] == t[h]) {
            m2[h - i] = max(m2[h - i], mx2[h - i - 1]);
            m1[mx2[h - i - 1]] = max(m1[mx2[h - i - 1]], h - i);
          }
        }
        for (int h = 0; h <= min(j, i) + 1; h++) {
          mx1[h] = max(mx1[h], m1[h]);
          m1[h] = -1;
        }
        for (int h = 0; h <= min(j, (int)s.size() - i - 2) + 1; h++) {
          mx2[h] = max(mx2[h], m2[h]);
          m2[h] = -1;
        }
      }
      if (mx1[i + 1] == (int)t.size() - (i + 1)) {
        flag = true;
        break;
      }
    }
    if (flag) {
      printf("YES\n");
    } else {
      printf("NO\n");
    }
  }
  return 0;
}
