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;
char a[102333], b[102333];
int A[102333], B[102333];
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    int n;
    scanf("%d", &n);
    scanf("%s%s", a + 1, b + 1);
    for (int i = n; i >= 1; --i) A[i] = a[i] - '0';
    for (int i = n; i >= 1; --i) B[i] = b[i] - '0';
    int L = 1, R = n, rev = 0;
    vector<int> V;
    for (int i = n; i >= 1; --i) {
      if (rev) {
        if ((A[L] ^ rev) == B[i]) {
          ++L;
          continue;
        }
      } else {
        if ((A[R] ^ rev) == B[i]) {
          --R;
          continue;
        }
      }
      if (rev) {
        if (A[R] ^ B[i] ^ rev)
          V.push_back(R - L + 1);
        else
          V.push_back(1), V.push_back(R - L + 1);
        --R;
      } else {
        if (A[L] ^ B[i] ^ rev)
          V.push_back(R - L + 1);
        else
          V.push_back(1), V.push_back(R - L + 1);
        ++L;
      }
      rev = !rev;
    }
    printf("%d", V.size());
    for (int u : V) printf(" %d", u);
    printf("\n");
  }
  return 0;
}