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.
#include <bits/stdc++.h>
using namespace std;
bool isrange(int second, int first, int n, int m) {
  if (0 <= second && second < n && 0 <= first && first < m) return true;
  return false;
}
int dy[4] = {1, 0, -1, 0}, dx[4] = {0, 1, 0, -1},
    ddy[8] = {1, 0, -1, 0, 1, 1, -1, -1}, ddx[8] = {0, 1, 0, -1, 1, -1, 1, -1};
void oper(string& a, int k) {
  string r = "";
  for (int e = 0; e <= k; e++) {
    if (a[e] == '1')
      r += "0";
    else
      r += "1";
  }
  reverse(r.begin(), r.end());
  for (int e = 0; e <= k; e++) a[e] = r[e];
}
int main(void) {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    string a, b;
    cin >> n >> a >> b;
    vector<int> ans;
    for (int e = n - 1; e >= 0; e--) {
      if (a[e] != b[e]) {
        if (a[0] != a[e]) {
          ans.push_back(1);
          oper(a, 0);
        }
        ans.push_back(e + 1);
        oper(a, e);
      }
    }
    printf("%d ", (int)ans.size());
    for (int e = 0; e < (int)ans.size(); e++) printf("%d ", ans[e]);
    printf("\n");
  }
  return 0;
}
