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.
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  if (a.first == b.first) return (a.second < b.second);
  return (a.first > b.first);
}
void dvpii(vector<pair<int, int>> v) {
  for (int i = (0); i < (v.size()); i++) {
    cout << i + 1 << " " << v[i].first << " " << v[i].second << "\n";
  }
}
int gcd(int a, int b) {
  if (a == 0) {
    return b;
  }
  return gcd(b % a, a);
}
bool isprime(int n) {
  if (n % 2 == 0 || n % 3 == 0) return false;
  int i = 5;
  while (i * i <= n) {
    if (n % (i) == 0 || n % (i + 2) == 0) return false;
    i += 6;
  }
  return true;
}
void canonical(int n, vector<pair<int, int>> &ans) {
  for (int i = 2; i * i <= n; i++) {
    int cnt = 0;
    while (n % i == 0) {
      cnt++;
      n /= i;
    }
    if (cnt > 0) {
      ans.push_back({i, cnt});
    }
  }
}
int divisors(int a) {
  int cnt = 1;
  for (int i = 2; i <= sqrt(a); i++) {
    if (a % i == 0) {
      if (a / i == i) {
        cnt++;
      } else {
        cnt += 2;
      }
    }
  }
  return cnt;
}
void solve() {
  int n;
  cin >> n;
  string a, b;
  cin >> a >> b;
  vector<int> ans;
  for (int i = (0); i < (n); i++) {
    if (a[i] != b[i]) {
      ans.push_back(i + 1);
      ans.push_back(1);
      ans.push_back(i + 1);
    }
  }
  cout << ans.size() << " ";
  for (int X : ans) cout << X << " ";
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
}
