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;
int arr[100007];
bool isPalRec(string str, int s, int e) {
  if (s == e) return true;
  if (str[s] != str[e]) return false;
  if (s < e + 1) return isPalRec(str, s + 1, e - 1);
  return true;
}
bool isPrime(int n) {
  if (n <= 1) return false;
  for (int i = 2; i < sqrt(n); i++)
    if (n % i == 0) return false;
  return true;
}
vector<long long int> primeFactors(long long int n) {
  vector<long long int> v;
  while (n % 2 == 0) {
    v.push_back(2);
    n = n / 2;
  }
  for (long long int i = 3; i <= sqrt(n); i = i + 2) {
    while (n % i == 0) {
      v.push_back(i);
      n = n / i;
    }
  }
  if (n > 2) v.push_back(n);
  return v;
}
bool solve() {
  long long int n;
  cin >> n;
  long long int a[n];
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long sum = 0;
  for (int i = 0; i < n; i++) {
    sum += a[i];
    if (sum <= 0) return 0;
  }
  sum = 0;
  for (int i = n - 1; i >= 0; i--) {
    sum += a[i];
    if (sum <= 0) return 0;
  }
  return 1;
}
int main() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int n, i, j;
    string a, b;
    cin >> n;
    cin >> a >> b;
    vector<int> ans;
    for (i = 0; i < n; i++) {
      if (a[i] != b[i]) {
        if (i > 0) ans.push_back(i + 1);
        ans.push_back(1);
        if (i > 0) ans.push_back(i + 1);
      }
    }
    cout << ans.size() << " ";
    for (auto it : ans) cout << it << " ";
    cout << endl;
  }
}