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;
int main() {
  long long n, i, x, j, t, y, k;
  cin >> t;
  while (t--) {
    long long m;
    cin >> n;
    string p, q;
    cin >> p >> q;
    long long r = 0;
    long long first = 0, end = n - 1;
    long long x = n;
    vector<long long> v;
    for (i = n - 1; i > 0; i--) {
      if (p[i] != q[i]) {
        if (p[0] == '0' && p[i] == '1') {
          for (j = 1; j <= i; j++) {
            if (p[j] == '1')
              p[j] = '0';
            else
              p[j] = '1';
          }
          k = i;
          j = 0;
          while (j < k) {
            char c = p[j];
            p[j] = p[k];
            p[k] = c;
            j++;
            k--;
          }
          v.push_back(1);
          v.push_back(i + 1);
        } else if (p[0] == '1' && p[i] == '1') {
          for (j = 0; j <= i; j++) {
            if (p[j] == '1')
              p[j] = '0';
            else
              p[j] = '1';
          }
          k = i;
          j = 0;
          while (j < k) {
            char c = p[j];
            p[j] = p[k];
            p[k] = c;
            j++;
            k--;
          }
          v.push_back(i + 1);
        } else if (p[0] == '0' && p[i] == '0') {
          for (j = 0; j <= i; j++) {
            if (p[j] == '1')
              p[j] = '0';
            else
              p[j] = '1';
          }
          k = i;
          j = 0;
          while (j < k) {
            char c = p[j];
            p[j] = p[k];
            p[k] = c;
            j++;
            k--;
          }
          v.push_back(i + 1);
        } else {
          for (j = 1; j <= i; j++) {
            if (p[j] == '1')
              p[j] = '0';
            else
              p[j] = '1';
          }
          k = i;
          j = 0;
          while (j < k) {
            char c = p[j];
            p[j] = p[k];
            p[k] = c;
            j++;
            k--;
          }
          v.push_back(1);
          v.push_back(i + 1);
        }
      }
    }
    if (p[0] != q[0]) {
      v.push_back(1);
    }
    cout << v.size() << " ";
    for (i = 0; i < v.size(); i++) {
      cout << v[i] << " ";
    }
    cout << endl;
  }
  return 0;
}