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>
const int MAXN = 400;
const int CS = 26;
const int INF = 0x3f3f3f3f;
using namespace std;
int dp[MAXN + 5][MAXN + 5], n, m;
char s[MAXN + 5], t[MAXN + 5];
int nxt[MAXN + 5][CS];
bool chk(int pos) {
  for (int i = 0; i <= pos; ++i) {
    for (int j = pos; j <= m; ++j) {
      dp[i][j] = INF;
    }
  }
  dp[0][pos] = 0;
  for (int i = 0; i <= pos; ++i) {
    for (int j = pos; j <= m; ++j) {
      if (dp[i][j] > n) continue;
      if (i < pos)
        dp[i + 1][j] = min(dp[i + 1][j], nxt[dp[i][j]][t[i + 1] - 'a']);
      if (j < m)
        dp[i][j + 1] = min(dp[i][j + 1], nxt[dp[i][j]][t[j + 1] - 'a']);
    }
  }
  return dp[pos][m] <= n;
}
int p[MAXN + 5];
void init() {
  scanf("%s%s", s + 1, t + 1);
  n = strlen(s + 1), m = strlen(t + 1);
  for (int c = 0; c < 26; ++c) p[c] = n + 1;
  for (int i = n; ~i; --i) {
    for (int c = 0; c < 26; ++c) nxt[i][c] = p[c];
    if (i) p[s[i] - 'a'] = i;
  }
}
int main() {
  int te;
  scanf("%d", &te);
  while (te--) {
    init();
    int cur = 1;
    for (int i = 1; i <= n; ++i) {
      if (cur > m) break;
      if (s[i] == t[cur]) ++cur;
    }
    if (cur > m) {
      puts("YES");
      continue;
    }
    bool f = false;
    for (int i = 1; i <= m; ++i) {
      if (chk(i)) {
        f = true;
        break;
      }
    }
    puts(f ? "YES" : "NO");
  }
  return 0;
}
