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 power(long long a, long long b) {
  long long res = 1;
  a = a % 998244353;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % 998244353;
      b--;
    }
    a = (a * a) % 998244353;
    b >>= 1;
  }
  return res;
}
long long fermat_inv(long long y) { return power(y, 998244353 - 2); }
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    string s;
    cin >> s;
    string s1;
    cin >> s1;
    int a[n], b[n];
    for (int i = 0; i < n; i++) {
      a[i] = s[i] - '0';
      b[i] = s1[i] - '0';
    }
    int ctr = 0;
    vector<int> ans;
    int c1 = 0;
    int c2 = n - 1;
    for (int i = n - 1; i >= 0; i--) {
      if (ctr % 2 == 0) {
        if ((a[c2] + ctr) % 2 == b[i])
          c2--;
        else {
          if ((a[c1] + ctr) % 2 == b[i]) {
            ans.push_back(1);
          }
          ans.push_back(c2 - c1 + 1);
          ctr++;
          c1++;
        }
      } else {
        if ((a[c1] + ctr) % 2 == b[i])
          c1++;
        else {
          if ((a[c2] + ctr) % 2 == b[i]) {
            ans.push_back(1);
          }
          ans.push_back(c2 - c1 + 1);
          ctr++;
          c2--;
        }
      }
    }
    cout << ans.size() << " ";
    for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
    cout << "\n";
  }
  return 0;
}