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 char gc() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline long long read() {
  long long x = 0;
  char ch = getchar();
  bool positive = 1;
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') positive = 0;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  return positive ? x : -x;
}
inline void write(long long x) {
  if (x < 0) {
    x = -x;
    putchar('-');
  }
  if (x >= 10) write(x / 10);
  putchar('0' + x % 10);
}
inline void writeln(long long x) {
  write(x);
  puts("");
}
inline void writep(long long x) {
  write(x);
  putchar(' ');
}
inline unsigned long long rnd() {
  return ((unsigned long long)rand() << 30 ^ rand()) << 4 | rand() % 4;
}
const int N = 405;
int f[N][N];
char s[N], t[N];
int main() {
  int T = read();
  while (T--) {
    memset(f, 0, sizeof(f));
    scanf("%s%s", s + 1, t + 1);
    int n = strlen(s + 1), m = strlen(t + 1);
    bool d = 1;
    for (int i = 1, pos = 1; i <= m; i++) {
      while (pos <= n && s[pos] != t[i]) pos++;
      if (pos > n) {
        d = 0;
        break;
      }
      pos++;
    }
    if (d) {
      puts("YES");
      continue;
    }
    for (int o = (int)(2); o <= (int)(m); o++) {
      memset(f, -1, sizeof(f));
      f[0][0] = 0;
      for (int i = (int)(0); i <= (int)(n); i++)
        for (int j = (int)(0); j <= (int)(o - 1); j++)
          if (f[i][j] != -1) {
            f[i + 1][j] = max(f[i][j], f[i + 1][j]);
            int pos = o + f[i][j], ssw = f[i][j];
            if (j < o - 1 && s[i + 1] == t[j + 1])
              f[i + 1][j + 1] = max(f[i + 1][j + 1], ssw);
            if (pos <= m && s[i + 1] == t[pos])
              f[i + 1][j] = max(f[i + 1][j], ssw + 1);
          }
      if (f[n][o - 1] == m - o + 1) {
        d = 1;
        break;
      }
    }
    if (d)
      puts("YES");
    else
      puts("NO");
  }
}
