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;
template <class T>
ostream& operator<<(ostream& cout, vector<T> V) {
  cout << "[ ";
  for (auto v : V) cout << v << ' ';
  return cout << ']';
}
template <class L, class R>
ostream& operator<<(ostream& cout, pair<L, R> P) {
  return cout << '(' << P.first << ',' << P.second << ')';
}
void invert(string& a, long long int ind) {
  for (long long int i = 0; i < (ind + 1); i++) {
    if (a[i] == '1')
      a[i] = '0';
    else
      a[i] = '1';
  }
  for (long long int i = 0, j = ind; i < j; i++, j--) swap(a[i], a[j]);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    string a, b;
    cin >> a >> b;
    string bac = a;
    vector<long long int> ans;
    long long int inv = 0, rev = 0;
    long long int l = 0, r = n - 1;
    for (long long int i = n - 1; i >= 0; i--) {
      if (rev) {
        if (inv) {
          if (a[l] == b[i]) {
            if (a[l] == a[r]) {
              ans.push_back(i);
              inv = 1 - inv;
              rev = 1 - rev;
            } else {
              if (a[r] == '0')
                a[r] = '1';
              else
                a[r] = '0';
              ans.push_back(0);
              ans.push_back(i);
              inv = 1 - inv;
              rev = 1 - rev;
            }
          }
        } else {
          if (a[l] != b[i]) {
            if (a[l] == a[r]) {
              ans.push_back(i);
              inv = 1 - inv;
              rev = 1 - rev;
            } else {
              if (a[r] == '0')
                a[r] = '1';
              else
                a[r] = '0';
              ans.push_back(0);
              ans.push_back(i);
              inv = 1 - inv;
              rev = 1 - rev;
            }
          }
        }
        if (rev)
          l++;
        else
          r--;
      } else {
        if (inv) {
          if (a[r] == b[i]) {
            if (a[l] == a[r]) {
              ans.push_back(i);
              inv = 1 - inv;
              rev = 1 - rev;
            } else {
              if (a[l] == '0')
                a[l] = '1';
              else
                a[l] = '0';
              ans.push_back(0);
              ans.push_back(i);
              inv = 1 - inv;
              rev = 1 - rev;
            }
          }
        } else {
          if (a[r] != b[i]) {
            if (a[l] == a[r]) {
              ans.push_back(i);
              inv = 1 - inv;
              rev = 1 - rev;
            } else {
              if (a[l] == '0')
                a[l] = '1';
              else
                a[l] = '0';
              ans.push_back(0);
              ans.push_back(i);
              inv = 1 - inv;
              rev = 1 - rev;
            }
          }
        }
        if (!rev)
          r--;
        else
          l++;
      }
    }
    cout << ans.size() << ' ';
    for (long long int i : ans) cout << i + 1 << ' ';
    cout << '\n';
  }
  cerr << "Time : " << 1000 * (long double)clock() / (long double)CLOCKS_PER_SEC
       << "ms\n";
  ;
}