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>
using namespace std;
long long cl(long long n, long long m) {
  if (n % m)
    return n / m + 1;
  else
    return n / m;
}
long long pw(long long a, long long b, long long M) {
  a %= M;
  long long r = 1;
  while (b > 0) {
    if (b & 1) r = r * a % M;
    a = a * a % M;
    b >>= 1;
  }
  return r;
}
const long long N = 1e6 + 5;
char mul(char t, long long x) {
  long long b = t - '0';
  b = b ^ x;
  char c = b + '0';
  return c;
}
signed main() {
  long long TT;
  cin >> TT;
  while (TT--) {
    long long n;
    cin >> n;
    string a, b;
    cin >> a >> b;
    long long x = 0;
    long long l = 0, r = a.size() - 1;
    long long t = b.size() - 1;
    vector<long long> ans;
    for (long long i = t; i >= 0; i--) {
      if (l <= r) {
        if (mul(a[r], x) == b[i]) {
          r--;
          continue;
        } else {
          if (a[l] == a[r]) {
            ans.push_back(i + 1);
            x = x ^ 1;
            l++;
            swap(l, r);
          } else {
            ans.push_back(1);
            a[l] = mul(a[l], 1);
            ans.push_back(i + 1);
            x = x ^ 1;
            l++;
            swap(l, r);
          }
        }
      } else {
        if (mul(a[r], x) == b[i]) {
          r++;
          continue;
        } else {
          if (a[l] == a[r]) {
            ans.push_back(i + 1);
            x = x ^ 1;
            l--;
            swap(l, r);
          } else {
            ans.push_back(1);
            a[l] = mul(a[l], 1);
            ans.push_back(i + 1);
            x = x ^ 1;
            l--;
            swap(l, r);
          }
        }
      }
    }
    cout << ans.size() << " ";
    for (auto it : ans) {
      cout << it << " ";
    }
    cout << endl;
  }
}
