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;
long long digits(long long n) {
  long long cnt = 0;
  while (n > 0) {
    cnt++;
    n = n / 10;
  }
  return cnt;
}
long long nck(long long n, long long k) {
  long long C[n + 1][k + 1];
  long long i, j;
  for (i = 0; i <= n; i++) {
    for (j = 0; j <= min(i, k); j++) {
      if (j == 0 || j == i)
        C[i][j] = 1;
      else
        C[i][j] = (C[i - 1][j - 1] % 1000000007 + C[i - 1][j] % 1000000007) %
                  1000000007;
    }
  }
  return C[n][k] % 1000000007;
}
long long modularExponentiation(long long x, long long n, long long M) {
  long long result = 1;
  while (n > 0) {
    if (n % 2 == 1) {
      result = ((result % M) * (x % M)) % M;
    }
    x = ((x % M) * (x % M)) % M;
    n = n / 2;
  }
  return result;
}
int32_t main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    string a, b;
    cin >> n >> a >> b;
    vector<long long> ope;
    for (long long i = 1; i < n; i++) {
      if (a[i] != a[i - 1]) ope.push_back(i);
    }
    if (a[n - 1] == '1') ope.push_back(n);
    char curr = '0';
    for (long long i = n - 1; i >= 0; i--) {
      if (b[i] != curr) {
        if (curr == '0')
          curr = '1';
        else
          curr = '0';
        ope.push_back(i + 1);
      }
    }
    cout << ope.size() << " ";
    for (long long i = 0; i < ope.size(); i++) cout << ope[i] << " ";
    cout << endl;
  }
}