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;
const int MX = 500005;
const long long MOD = 1000000007;
void setIO(string name) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  freopen((name + ".in").c_str(), "r", stdin);
  freopen((name + ".out").c_str(), "w", stdout);
}
int t;
int n;
bool need[MX];
void solve() {
  cin >> n;
  string a, b;
  cin >> a;
  cin >> b;
  int tot = 0;
  for (int i = 0; i < (n); i++) {
    if (i % 2 == 0) {
      if (a[i / 2] == b[n - 1 - i]) {
        need[i] = 1;
        tot++;
      } else
        need[i] = 0;
    } else {
      if (a[n - 1 - i / 2] != b[n - 1 - i]) {
        need[i] = 1;
        tot++;
      } else
        need[i] = 0;
    }
  }
  cout << n + tot;
  for (int i = 0; i < (n); i++) {
    if (need[i]) cout << ' ' << 1;
    cout << ' ' << n - i;
  }
  cout << endl;
}
int main() {
  cin >> t;
  for (int i = 0; i < (t); i++) {
    solve();
  }
}