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 linf = (1ll << 62) - 1;
const int N = 1e3 + 7;
const int mod = 1e9 + 7;
const double eps = 1e-12;
template <typename T>
inline void rd(T &x) {
  char c;
  int sign = 1;
  x = 0;
  do {
    c = getchar();
    if (c == '-') sign = -1;
  } while (c < '0' || c > '9');
  do {
    x = x * 10 + c - '0';
    c = getchar();
  } while (c >= '0' && c <= '9');
  x *= sign;
}
int nx[N][500], f[N][N];
char s[N], t[N], s1[N], s2[N];
int T, lent, len;
void get() {
  memset(nx, 0, sizeof(nx));
  for (int i = 0; i <= 25; ++i) nx[len][i] = len + 1;
  for (int i = len; i >= 1; --i) {
    for (int j = 0; j <= 25; ++j) nx[i - 1][j] = nx[i][j];
    nx[i - 1][s[i] - 'a'] = i;
  }
}
void dp(int n1) {
  for (int i = 0; i <= lent; ++i)
    for (int j = 0; j <= lent; ++j) f[i][j] = inf;
  f[0][0] = 0;
  for (int i = 1; i <= n1; ++i) s1[i] = t[i];
  for (int i = n1 + 1; i <= lent; ++i) s2[i - n1] = t[i];
  for (int i = 0; i <= n1; ++i) {
    for (int j = 0; j <= lent - n1; ++j) {
      if (!i && !j) continue;
      if (i && f[i - 1][j] != inf && nx[f[i - 1][j]][s1[i] - 'a'])
        f[i][j] = min(f[i][j], nx[f[i - 1][j]][s1[i] - 'a']);
      if (j && f[i][j - 1] != inf && nx[f[i][j - 1]][s2[j] - 'a'])
        f[i][j] = min(f[i][j], nx[f[i][j - 1]][s2[j] - 'a']);
    }
  }
}
int main() {
  scanf("%d", &T);
  while (T--) {
    int flag = 0;
    scanf("%s%s", s + 1, t + 1);
    lent = strlen(t + 1), len = strlen(s + 1);
    get();
    for (int i = 1; i <= lent; ++i) {
      dp(i);
      if (f[i][lent - i] <= len) {
        flag = 1;
        break;
      }
    }
    puts(flag ? "YES" : "NO");
  }
  return 0;
}
