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("Ofast,no-stack-protector,unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int t = 1;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    string a, b;
    cin >> a >> b;
    long long c = 0, inv = 0;
    vector<long long> v;
    map<char, char> m;
    m['1'] = '0';
    m['0'] = '1';
    long long l = 0, r = n - 1;
    for (long long i = n - 1; i > 0; i--) {
      if (inv == 0) {
        bool flag = false;
        if (a[r] != b[i]) {
          flag = true;
          if (a[l] == b[i]) {
            c++;
            v.push_back(1);
            a[l] = m[a[l]];
          }
          c++;
          v.push_back(i + 1);
          inv = 1 - inv;
          swap(l, r);
          r++;
        }
        if (!flag) r--;
      } else {
        bool flag = false;
        if (a[r] == b[i]) {
          flag = true;
          if (a[l] != b[i]) {
            c++;
            v.push_back(1);
            a[l] = m[a[l]];
          }
          c++;
          v.push_back(i + 1);
          inv = 1 - inv;
          swap(l, r);
          r--;
        }
        if (!flag) r++;
      }
    }
    if (inv == 0) {
      if (a[r] != b[0]) {
        c++;
        v.push_back(1);
      }
    } else {
      if (a[r] == b[0]) {
        c++;
        v.push_back(1);
      }
    }
    cout << c << " ";
    for (auto i : v) cout << i << " ";
    cout << "\n";
    ;
  }
  return 0;
}
