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 = 404;
const short inf = 2000;
char s[N], t[N];
short dp[N][N][N], vis[N][N][N], cn, ls, lt;
short rec(int p, int i, int j) {
  if (p == ls) {
    if (i == lt || j == lt) return 0;
    return -inf;
  }
  short &ret = dp[p][i][j];
  short &vs = vis[p][i][j];
  if (vs == cn) return ret;
  vs = cn;
  ret = -inf;
  int x = 0;
  if (i < lt && s[p] == t[i])
    x = 1, ret = max(ret, (short)(rec(p + 1, i + 1, j) + 1));
  if (j < lt && s[p] == t[j])
    x = 1, ret = max(ret, (short)(rec(p + 1, i, j + 1) + 1));
  if (!x) ret = rec(p + 1, i, j);
  return ret;
}
int main() {
  int tc;
  scanf("%d", &tc);
  while (tc--) {
    scanf("%s%s", s, t);
    ls = strlen(s);
    lt = strlen(t);
    int mx = 0;
    cn++;
    for (int i = 0; i < lt; i++) mx = max(mx, (int)rec(0, 0, i));
    if (mx >= lt)
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}
