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 m = 1000000007;
long long fact(long long n);
int modInverse(int a, int m) {
  int m0 = m;
  int y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    int q = a / m;
    int t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += m0;
  return x;
}
long long fact(long long n) {
  long long res = 1;
  for (long long i = 2; i <= n; i++) res = (res % m * i % m) % m;
  return res;
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, an = 0, st = 0, en, c = 0;
    cin >> n;
    string a, b;
    cin >> a >> b;
    en = n - 1;
    vector<long long> v;
    for (long long i = 0; i < n; i++) {
      if (an % 2 == 0) {
        if ((v.size() - c) % 2 == 1) {
          if (a[st] == '0')
            a[st]++;
          else
            a[st]--;
        }
        if (a[st] == b[n - i - 1]) {
          v.push_back(1);
          c++;
          v.push_back(n - i);
        } else {
          v.push_back(n - i);
        }
        st++;
        an++;
      } else {
        if ((v.size() - c) % 2 == 1) {
          if (a[en] == '0')
            a[en]++;
          else
            a[en]--;
        }
        if (a[en] == b[n - i - 1]) {
          v.push_back(1);
          c++;
          v.push_back(n - i);
        } else {
          v.push_back(n - i);
        }
        en--;
        an++;
      }
    }
    cout << v.size() << " ";
    for (long long i = 0; i < v.size(); i++) cout << v[i] << " ";
    cout << endl;
  }
  return 0;
}