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 INF = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-9;
inline int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
inline int lcm(int a, int b) { return a * b / gcd(a, b); }
inline int rd() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
const int maxn = 4e2 + 50;
string s, s1;
int t[maxn][26], tt, dp[maxn][maxn];
void SEAM() {
  for (int i = s.size() - 1; i >= 0; i--) {
    for (int j = 0; j <= 25; j++) t[i][j] = t[i + 1][j];
    t[i][s[i] - 'a'] = i;
  }
}
bool check(string a, string b) {
  memset(dp, inf, sizeof dp);
  dp[0][0] = 0;
  for (int i = 0; i <= a.size(); i++) {
    for (int j = 0; j <= b.size(); j++) {
      int pos = dp[i][j];
      if (pos > s.size()) continue;
      if (i < a.size() && t[pos][a[i] - 'a'] < inf) {
        dp[i + 1][j] = min(dp[i + 1][j], t[pos][a[i] - 'a'] + 1);
      }
      if (j < b.size() && t[pos][b[j] - 'a'] < inf) {
        dp[i][j + 1] = min(dp[i][j + 1], t[pos][b[j] - 'a'] + 1);
      }
    }
  }
  return dp[a.size()][b.size()] < inf;
}
void run() {
  cin >> tt;
  while (tt--) {
    cin >> s >> s1;
    memset(t, inf, sizeof t);
    SEAM();
    int ff = 0;
    for (int i = 0; i <= s1.size() - 1; i++) {
      if (check(s1.substr(0, i + 1), s1.substr(i + 1, s1.size()))) {
        ff = 1;
        break;
      }
    }
    if (ff == 1)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
}
int main() {
  run();
  return 0;
}
