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>
const int N = (int)5e5 + 7;
const int inf = (int)1e9 + 7;
const int mod = (int)1e9 + 7;
const long long linf = (long long)1e18 + 7;
const int dx[] = {-1, 0, 1, 0, 1, -1, -1, 1};
const int dy[] = {0, 1, 0, -1, 1, -1, 1, -1};
using namespace std;
int n;
int a[N], b[N];
void go(int l, int r) {
  for (int i = (l); i <= (r); i++) a[i] ^= 1;
  while (l < r) {
    swap(a[l], a[r]);
    l++, r--;
  }
}
void solve() {
  cin >> n;
  for (int i = (1); i <= (n); i++) {
    char c;
    cin >> c;
    a[i] = c - '0';
  }
  for (int i = (1); i <= (n); i++) {
    char c;
    cin >> c;
    b[i] = c - '0';
  }
  int first = 0;
  vector<int> ans;
  for (int i = (n); i >= (1); i--) {
    if (a[i] != b[i]) {
      if ((a[1] ^ 1) == b[i]) {
        go(1, i);
        ans.push_back(i);
        ;
      } else {
        go(1, 1);
        go(1, i);
        ans.push_back(1);
        ans.push_back(i);
      }
    }
  }
  cout << (int)ans.size() << ' ';
  for (auto it : ans) cout << it << ' ';
  cout << '\n';
}
int main() {
  ios_base ::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  exit(0);
}
