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;
inline int read() {
  char ch = getchar();
  int nega = 1;
  while (!isdigit(ch)) {
    if (ch == '-') nega = -1;
    ch = getchar();
  }
  int ans = 0;
  while (isdigit(ch)) {
    ans = ans * 10 + ch - 48;
    ch = getchar();
  }
  if (nega == -1) return -ans;
  return ans;
}
int add(int x, int y) { return x + y >= 998244353 ? x + y - 998244353 : x + y; }
int sub(int x, int y) { return x - y < 0 ? x - y + 998244353 : x - y; }
int mul(int x, int y) { return 1LL * x * y % 998244353; }
int f[405][405], n, m;
char s[405], t[405];
char a[405], b[405];
int nxt[405][405];
int la, lb;
bool judge(int x) {
  la = x, lb = m - x;
  for (int i = 1; i <= x; i++) a[i] = t[i];
  for (int i = x + 1; i <= m; i++) b[i - x] = t[i];
  for (int i = 1; i <= la; i++) f[i][0] = nxt[f[i - 1][0]][a[i] - 'a'];
  for (int i = 1; i <= lb; i++) f[0][i] = nxt[f[0][i - 1]][b[i] - 'a'];
  for (int i = 1; i <= la; i++) {
    for (int j = 1; j <= lb; j++) {
      f[i][j] = min(nxt[f[i - 1][j]][a[i] - 'a'], nxt[f[i][j - 1]][b[j] - 'a']);
    }
  }
  return f[la][lb] <= n;
}
void work() {
  scanf("%s%s", s + 1, t + 1);
  n = strlen(s + 1), m = strlen(t + 1);
  if (m == 1) {
    for (int i = 1; i <= n; i++) {
      if (s[i] == t[1]) {
        cout << "YES\n";
        return;
      }
    }
    cout << "NO\n";
    return;
  }
  for (int j = 0; j < 26; j++) nxt[n + 1][j] = n + 1;
  for (int i = n; i >= 0; i--) {
    for (int j = 0; j < 26; j++) {
      if (s[i + 1] - 'a' == j)
        nxt[i][j] = i + 1;
      else
        nxt[i][j] = nxt[i + 1][j];
    }
  }
  for (int i = 1; i <= m - 1; i++) {
    if (judge(i)) {
      cout << "YES\n";
      return;
    }
  }
  cout << "NO\n";
}
signed main() {
  int T = read();
  while (T--) work();
  return 0;
}
