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 maxn = 400 + 7;
int dp[maxn][maxn], lens, lent, fla;
char s[maxn], t[maxn];
int sol(int a, int b, int c, int d) {
  int len1 = b - a + 1;
  int len2 = d - c + 1;
  dp[0][0] = 0;
  for (int i = 1; i <= len1; i++) dp[0][i] = -1;
  for (int i = 1; i <= lens; i++) {
    dp[i][0] = dp[i - 1][0];
    if (s[i] == t[c - 1 + dp[i - 1][0] + 1]) dp[i][0]++;
    for (int j = 1; j <= len1; j++) {
      dp[i][j] = dp[i - 1][j];
      if (s[i] == t[a - 1 + j] && dp[i - 1][j - 1] != -1)
        dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
      if (dp[i - 1][j] != -1 && s[i] == t[c - 1 + dp[i - 1][j] + 1])
        dp[i][j] = max(dp[i][j], dp[i - 1][j] + 1);
    }
  }
  if (dp[lens][len1] == len2) {
    return 1;
  } else
    return 0;
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    fla = 0;
    scanf("%s", s + 1);
    scanf("%s", t + 1);
    lens = strlen(s + 1);
    lent = strlen(t + 1);
    int cnt = 0;
    for (int i = 1; i <= lens; i++)
      if (cnt + 1 <= lent && t[cnt + 1] == s[i]) cnt++;
    if (cnt == lent) {
      printf("YES\n");
      continue;
    }
    for (int i = 1; i < lent; i++) fla += sol(1, i, i + 1, lent);
    if (fla)
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}
