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 maxs = 405;
const int inf = 1e9 + 7;
int dp[maxs][maxs];
string s, t, x, y;
int slen, tlen;
int nxt[maxs][26];
int g(int x) { return x + 1; }
int f(int a, int b) {
  if (a == -1 && b == -1) return 0;
  if (a < -1 || b < -1) return inf;
  int& r = dp[g(a)][g(b)];
  if (r == -1) {
    r = inf;
    int idx;
    if (a >= 0) {
      idx = f(a - 1, b);
      if (idx < inf) r = min(r, nxt[idx][x[a] - 'a'] + 1);
    }
    if (b >= 0) {
      idx = f(a, b - 1);
      if (idx < inf) r = min(r, nxt[idx][y[b] - 'a'] + 1);
    }
  }
  return r;
}
int main(void) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    cin >> s >> t;
    int i = 0, j = 0;
    slen = s.size(), tlen = t.size();
    while (i < slen && j < tlen) {
      if (s[i] == t[j]) ++j;
      ++i;
    }
    if (j == tlen) {
      cout << "YES\n";
      continue;
    }
    for (int i = 0; i <= slen; ++i)
      for (int j = 0; j < 26; ++j) nxt[i][j] = inf;
    for (int i = slen - 1; i >= 0; --i) {
      for (int j = 0; j < 26; ++j) nxt[i][j] = nxt[i + 1][j];
      nxt[i][s[i] - 'a'] = i;
    }
    bool yes = false;
    for (int i = 1; i < tlen; ++i) {
      for (int j = 0; j <= tlen; ++j)
        for (int k = 0; k <= tlen; ++k) dp[j][k] = -1;
      x = t.substr(0, i);
      y = t.substr(i, tlen - i);
      if (f(i - 1, tlen - i - 1) < inf) {
        yes = true;
        break;
      }
    }
    cout << (yes ? "YES\n" : "NO\n");
  }
  return 0;
}
