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 inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3fll;
const int N = 400 + 10;
const int M = 100 + 10;
const int mod = 1000000009;
const double eps = 1e-10;
const double pi = acos(-1.0);
int n, m;
char s[N], a[N], b[N];
int nxt[N][30];
int dp[N][N];
inline int idx(char c) { return c - 'a'; }
bool check(int x) {
  int y = m - x;
  for (int i = 1; i <= y; i++) b[i] = a[i + x];
  memset(dp, -1, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 0; i <= x; i++) {
    for (int j = 0; j <= y; j++) {
      if (~dp[i][j]) {
        int pos = dp[i][j];
        if (i < x && ~nxt[pos][idx(a[i + 1])]) {
          if (dp[i + 1][j] == -1)
            dp[i + 1][j] = nxt[pos][idx(a[i + 1])];
          else
            dp[i + 1][j] = min(dp[i + 1][j], nxt[pos][idx(a[i + 1])]);
        }
        if (j < y && ~nxt[pos][idx(b[j + 1])]) {
          if (dp[i][j + 1] == -1)
            dp[i][j + 1] = nxt[pos][idx(b[j + 1])];
          else
            dp[i][j + 1] = min(dp[i][j + 1], nxt[pos][idx(b[j + 1])]);
        }
      }
    }
  }
  return dp[x][y] != -1;
}
void init() {
  memset(nxt[0], -1, sizeof(nxt[0]));
  for (int i = n; i; i--) {
    memcpy(nxt[i], nxt[0], sizeof(nxt[0]));
    nxt[0][idx(s[i])] = i;
  }
}
void work() {
  cin >> s + 1 >> a + 1;
  n = strlen(s + 1);
  m = strlen(a + 1);
  init();
  for (int i = 1; i <= m; i++) {
    if (check(i)) {
      cout << "YES\n";
      return;
    }
  }
  cout << "NO\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) work();
  return 0;
}
