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>
long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
using namespace std;
const long long maxn = 1e3 + 7;
const double eps = 1e-10;
const long long mod = 1e9 + 7;
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    string a, b;
    cin >> a >> b;
    long long now = n - 1;
    long long sum = 0;
    vector<long long> ans;
    long long L = 0, R = n - 1;
    while (now >= 0) {
      if (now == 0) {
        if (sum & 1) {
          if (a[L] == b[0]) ans.push_back(1);
        } else if (a[L] != b[0])
          ans.push_back(1);
      } else {
        long long l, r;
        if (sum & 1) {
          l = !(a[R] - '0');
          r = !(a[L] - '0');
          if (r == b[now] - '0') {
            L++;
          } else {
            R--;
            if (l == b[now] - '0') {
              ans.push_back(1);
              ans.push_back(now + 1);
              sum++;
            } else {
              ans.push_back(now + 1);
              sum++;
            }
          }
        } else {
          l = a[L] - '0';
          r = a[R] - '0';
          if (r == b[now] - '0') {
            R--;
          } else {
            L++;
            if (l == b[now] - '0') {
              ans.push_back(1);
              ans.push_back(now + 1);
              sum++;
            } else {
              ans.push_back(now + 1);
              sum++;
            }
          }
        }
      }
      now--;
    }
    cout << (long long)ans.size();
    for (long long i = 0; i < ans.size(); i++) cout << ' ' << ans[i];
    cout << "\n";
  }
}