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>
#pragma warning(disable : 4996)
using namespace std;
const int maxn = 1e3 + 5;
queue<int> Q;
char s1[maxn], s2[maxn];
int main() {
  int t, n;
  cin >> t;
  while (t--) {
    cin >> n;
    cin >> (s1 + 1);
    cin >> (s2 + 1);
    int op = 0;
    for (int i = (1); i <= (n); ++i) {
      if (s1[i] == '1') {
        int temp = i;
        while (s1[i + 1] == '1' && i <= n) {
          s1[i] = '0';
          i++;
        }
        if (temp == 1) {
          op++;
          Q.push(i);
          s1[i] = '0';
        } else {
          op += 2;
          Q.push(temp - 1);
          Q.push(i);
          s1[i] = '0';
        }
      }
    }
    for (int i = n; i >= 1; i--) {
      if (s2[i] == '1') {
        int temp = i;
        while (s2[i - 1] == '1' && i >= 1) {
          i--;
        }
        if (i == 1) {
          op++;
          Q.push(temp);
        } else {
          op += 2;
          Q.push(temp);
          Q.push(i - 1);
        }
      }
    }
    printf("%d", op);
    while (!Q.empty()) {
      int x = Q.front();
      Q.pop();
      printf(" %d", x);
    }
    printf("\n");
  }
  return 0;
}