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>
#pragma GCC optimize("Ofast")
using namespace std;
inline void rd(int &x) {
  int k;
  int m = 0;
  x = 0;
  for (;;) {
    k = getchar();
    if (k == '-') {
      m = 1;
      break;
    }
    if ('0' <= k && k <= '9') {
      x = k - '0';
      break;
    }
  }
  for (;;) {
    k = getchar();
    if (k < '0' || k > '9') {
      break;
    }
    x = x * 10 + k - '0';
  }
  if (m) {
    x = -x;
  }
}
inline void rd(char &c) {
  int i;
  for (;;) {
    i = getchar();
    if (i != ' ' && i != '\n' && i != '\r' && i != '\t' && i != EOF) {
      break;
    }
  }
  c = i;
}
inline int rd(char c[]) {
  int i;
  int sz = 0;
  for (;;) {
    i = getchar();
    if (i != ' ' && i != '\n' && i != '\r' && i != '\t' && i != EOF) {
      break;
    }
  }
  c[sz++] = i;
  for (;;) {
    i = getchar();
    if (i == ' ' || i == '\n' || i == '\r' || i == '\t' || i == EOF) {
      break;
    }
    c[sz++] = i;
  }
  c[sz] = '\0';
  return sz;
}
inline int rd_int(void) {
  int x;
  rd(x);
  return x;
}
inline void wt_L(char a) { putchar(a); }
inline void wt_L(const char c[]) {
  int i = 0;
  for (i = 0; c[i] != '\0'; i++) {
    putchar(c[i]);
  }
}
template <class S, class T>
inline S chmin(S &a, T b) {
  if (a > b) {
    a = b;
  }
  return a;
}
int Slen;
int Tlen;
char S[402];
char T[402];
int nx[26][402];
int dp[402][402];
int main() {
  int Lj4PdHRW;
  int KL2GvlyY = rd_int();
  for (Lj4PdHRW = (0); Lj4PdHRW < (KL2GvlyY); Lj4PdHRW++) {
    int i, j;
    int f1;
    int f2;
    Slen = rd(S);
    Tlen = rd(T);
    for (i = (0); i < (Slen); i++) {
      S[i] -= 'a';
    }
    for (i = (0); i < (Tlen); i++) {
      T[i] -= 'a';
    }
    for (j = (0); j < (26); j++) {
      nx[j][Slen] = 1073709056;
    }
    for (i = (Slen)-1; i >= (0); i--) {
      for (j = (0); j < (26); j++) {
        nx[j][i] = nx[j][i + 1];
      }
      nx[S[i]][i] = i;
    }
    for (f1 = (0); f1 < (Tlen); f1++) {
      f2 = Tlen - f1;
      for (i = (0); i < (f1 + 1); i++) {
        for (j = (0); j < (f2 + 1); j++) {
          dp[i][j] = 1073709056;
        }
      }
      dp[0][0] = 0;
      for (i = (0); i < (f1 + 1); i++) {
        for (j = (0); j < (f2 + 1); j++) {
          if (dp[i][j] >= Slen) {
            continue;
          }
          if (i < f1) {
            chmin(dp[i + 1][j], nx[T[i]][dp[i][j]] + 1);
          }
          if (j < f2) {
            chmin(dp[i][j + 1], nx[T[f1 + j]][dp[i][j]] + 1);
          }
        }
      }
      if (dp[f1][f2] < 1073709056) {
        wt_L("YES");
        wt_L('\n');
        goto Q5VJL1cS;
      }
    }
    wt_L("NO");
    wt_L('\n');
  Q5VJL1cS:;
  }
  return 0;
}
