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;
void read(int &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -f;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  x *= f;
}
void print(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (!x) return;
  print(x / 10), putchar(x % 10 + 48);
}
void write(int x) {
  if (!x)
    putchar('0');
  else
    print(x);
  putchar('\n');
}
const int maxn = 400 + 10;
const int inf = 1e9;
const double eps = 1e-8;
const int mod = 1e9 + 7;
char s[maxn], t[maxn];
int f[maxn][maxn], nxt[maxn][26];
int n, m;
void update(int &x, int y) {
  if (y < x) x = y;
}
int check(int r) {
  for (int i = 0; i <= r; i++)
    for (int j = 0; j <= m - r; j++) f[i][j] = inf;
  f[0][0] = 0;
  for (int i = 0; i <= r; i++)
    for (int j = 0; j <= m - r; j++) {
      if (f[i][j] == inf) continue;
      if (i != r) update(f[i + 1][j], nxt[f[i][j]][t[i + 1] - 'a']);
      if (j != m - r) update(f[i][j + 1], nxt[f[i][j]][t[r + j + 1] - 'a']);
    }
  return f[r][m - r] != inf;
}
void solve() {
  scanf("%s%s", s + 1, t + 1);
  n = strlen(s + 1), m = strlen(t + 1);
  for (int i = 0; i < 26; i++) nxt[n][i] = inf;
  for (int i = n - 1; ~i; i--)
    for (int j = 0; j < 26; j++)
      if (s[i + 1] == j + 'a')
        nxt[i][j] = i + 1;
      else
        nxt[i][j] = nxt[i + 1][j];
  for (int i = 0; i <= m; i++)
    if (check(i)) return puts("YES"), void();
  puts("NO");
}
int main() {
  int t;
  read(t);
  while (t--) solve();
  return 0;
}
