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>
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,abm,mmx,avx,avx2,popcnt,fma")
#pragma GCC optimize("fast-math")
using namespace std;
const long long mod = 1e9 + 7, mod2 = 998244353;
const long double eps = 1e-6;
const long long inf = 1e18;
const int base = 1e9;
const long double PI = 3.1415926;
void deb(long double first){};
int dp[401][401], T, prv[30];
string s, t;
unsigned int bitcount(unsigned long long val) {
  return __builtin_popcount(val) + __builtin_popcount(val >> 32);
}
void local() { freopen("test.in", "r", stdin); }
void remote() {}
void read() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cin >> T;
}
int main() {
  remote();
  read();
  while (T--) {
    memset(prv, -1, sizeof(prv));
    cin >> s >> t;
    vector<vector<int>> nxt(s.size() + 1, vector<int>(26, 500));
    for (int i = s.size() - 1; i >= 0; --i) {
      nxt[i] = nxt[i + 1];
      nxt[i][s[i] - 'a'] = i;
    }
    bool ex = false;
    for (int len = 0; len < t.size(); len++) {
      for (int i = 0; i < 401; i++)
        for (int j = 0; j < 401; j++) dp[i][j] = 500;
      dp[0][0] = 0;
      int a = len, b = t.size() - len;
      for (int i = 0; i < a + 1; i++)
        for (int j = 0; j < b + 1; j++) {
          if (dp[i][j] >= 500) continue;
          if (i < a && nxt[dp[i][j]][t[i] - 'a'] < 500) {
            dp[i + 1][j] = min(dp[i + 1][j], nxt[dp[i][j]][t[i] - 'a'] + 1);
          }
          if (j < b && nxt[dp[i][j]][t[len + j] - 'a'] < 500) {
            dp[i][j + 1] =
                min(dp[i][j + 1], nxt[dp[i][j]][t[len + j] - 'a'] + 1);
          }
        }
      if (dp[a][b] < 500) {
        ex = true;
        break;
      }
    }
    if (ex) {
      cout << "yes" << '\n';
      ;
    } else {
      cout << "no" << '\n';
      ;
    }
  }
}
