Problem: This is the easy version of the problem. The difference between the versions is the constraint on n and the required number of operations. You can make hacks only if all versions of the problem are solved.

There are two binary strings a and b of length n (a binary string is a string consisting of symbols 0 and 1). In an operation, you select a prefix of a, and simultaneously invert the bits in the prefix (0 changes to 1 and 1 changes to 0) and reverse the order of the bits in the prefix.

For example, if a=001011 and you select the prefix of length 3, it becomes 011011. Then if you select the entire string, it becomes 001001.

Your task is to transform the string a into b in at most 3n operations. It can be proved that it is always possible.

Input

The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 3t lines contain descriptions of test cases.

The first line of each test case contains a single integer n (1≤ n≤ 1000) — the length of the binary strings.

The next two lines contain two binary strings a and b of length n.

It is guaranteed that the sum of n across all test cases does not exceed 1000.

Output

For each test case, output an integer k (0≤ k≤ 3n), followed by k integers p_1,…,p_k (1≤ p_i≤ n). Here k is the number of operations you use and p_i is the length of the prefix you flip in the i-th operation.

Example

Input


5
2
01
10
5
01011
11100
2
01
01
10
0110011011
1000110100
1
0
1


Output


3 1 2 1
6 5 2 5 3 1 2
0
9 4 1 2 10 4 1 2 1 5
1 1

Note

In the first test case, we have 01→ 11→ 00→ 10.

In the second test case, we have 01011→ 00101→ 11101→ 01000→ 10100→ 00100→ 11100.

In the third test case, the strings are already the same. Another solution is to flip the prefix of length 2, which will leave a unchanged.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline int re_ad() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + (ch ^ 48), ch = getchar();
  return x * f;
}
inline bool read() {
  char ch = getchar();
  while (ch != '0' && ch != '1') ch = getchar();
  return (ch == '1');
}
int t, n;
bool A[100010], B[100010];
int cz[300010], cnt;
int to[300010];
int main() {
  register int i, j, fz, l, r, c;
  t = re_ad();
  while (t--) {
    n = re_ad();
    fz = 0;
    cnt = 0;
    l = 1;
    r = n;
    for (i = 1; i <= n; ++i) A[i] = read();
    for (i = 1; i <= n; ++i) B[i] = read();
    for (i = n; i > 1; --i) {
      if (fz)
        c = A[l] ^ fz;
      else
        c = A[r];
      if (c == B[i]) {
        if (fz)
          ++l;
        else
          --r;
        continue;
      }
      if (fz) {
        if ((A[r] ^ fz ^ 1) != B[i]) A[r] ^= 1, cz[++cnt] = 1;
      } else if ((A[l] ^ fz ^ 1) != B[i])
        A[l] ^= 1, cz[++cnt] = 1;
      cz[++cnt] = i;
      fz ^= 1;
      if (!fz)
        --r;
      else
        ++l;
    }
    A[l] ^= fz;
    if (A[l] != B[1]) cz[++cnt] = 1;
    printf("%d ", cnt);
    for (i = 1; i <= cnt; ++i) printf("%d ", cz[i]);
    putchar('\n');
  }
  return 0;
}