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 = 530;
const int M = 2 * N;
const long long MOD = 19260817LL;
inline long long lowbit(long long x) { return x & (-x); }
struct graph {
  int head[N], next[M], to[M];
  int tot;
  inline void init(int n) {
    for (int i = 0; i <= n; ++i) head[i] = -1;
    tot = 0;
  }
  inline void addedge(int x, int y) {
    to[tot] = y;
    next[tot] = head[x];
    head[x] = tot;
    ++tot;
  }
};
long long phi(long long x) {
  long long res = x;
  for (long long i = 2; i * i <= x; ++i) {
    if (x % i == 0) {
      res = res / i * (i - 1);
      while (x % i == 0) x /= i;
    }
  }
  if (x > 1) res = res / x * (x - 1);
  return res;
}
long long n, m;
char s[N], t[N];
int f[N][N], g[N][27];
int main() {
  int cases = 1;
  scanf("%d", &cases);
  for (int iii = 1; iii <= cases; ++iii) {
    scanf("%s", s + 1);
    scanf("%s", t + 1);
    n = strlen(s + 1);
    m = strlen(t + 1);
    for (int i = 0; i < 26; ++i) g[n + 1][i] = n + 1;
    for (int i = n; i >= 1; --i) {
      for (int j = 0; j < 26; ++j) g[i][j] = g[i + 1][j];
      g[i][s[i] - 'a'] = i;
    }
    bool ans = false;
    for (int i = 1; i <= m; ++i) {
      f[0][0] = 0;
      for (int j = 0; j <= i; ++j)
        for (int k = 0; k <= (m - i); ++k) {
          if (!j && !k) continue;
          f[j][k] = n + 1;
          if (j && f[j - 1][k] != n + 1) {
            int x = f[j - 1][k] + 1;
            int y = t[j] - 'a';
            f[j][k] = min(f[j][k], g[x][y]);
          }
          if (k && f[j][k - 1] != n + 1) {
            int x = f[j][k - 1] + 1;
            int y = t[k + i] - 'a';
            f[j][k] = min(f[j][k], g[x][y]);
          }
        }
      if (f[i][m - i] != n + 1) {
        ans = true;
        break;
      }
    }
    if (ans)
      puts("YES");
    else
      puts("NO");
  }
}
