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;
template <typename T>
void read(T &x) {
  x = 0;
  bool f = 0;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = 1;
  for (; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
  if (f) x = -x;
}
template <typename F>
inline void write(F x, char ed = '\n') {
  static short st[30];
  short tp = 0;
  if (x < 0) putchar('-'), x = -x;
  do st[++tp] = x % 10, x /= 10;
  while (x);
  while (tp) putchar('0' | st[tp--]);
  putchar(ed);
}
template <typename T>
inline void Mx(T &x, T y) {
  x < y && (x = y);
}
template <typename T>
inline void Mn(T &x, T y) {
  x > y && (x = y);
}
const int N = 405;
char s[N], t[N];
int f[N], T, fl;
bool work(int m, int n, int p) {
  int s1 = p, s2 = n - p;
  memset(f, -0x3f, sizeof(f));
  f[0] = 0;
  for (int i = 1; i <= m; i++) {
    for (int j = min(s1, i - 1); j >= 0; j--) {
      if (s[i] == t[j + 1]) Mx(f[j + 1], f[j]);
      if (f[j] >= 0 && f[j] < s2 && s[i] == t[f[j] + 1 + s1]) f[j]++;
    }
  }
  for (int i = 0; i <= s1; i++)
    if (f[i] + i >= n) return 1;
  return 0;
}
int main() {
  for (read(T); T; T--) {
    scanf("%s", s + 1);
    scanf("%s", t + 1);
    int n = strlen(s + 1);
    int m = strlen(t + 1);
    fl = 0;
    for (int i = 1; i <= m; i++)
      if (work(n, m, i)) {
        fl = 1;
        break;
      }
    puts(fl ? "YES" : "NO");
  }
  return 0;
}
