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;
mt19937 rnd(time(0));
bool isPrime(long long a) {
  long long i;
  for (i = 2; i * i <= a; i++) {
    if (a % i == 0) {
      return 0;
    }
  }
  return (a != 1);
}
bool isPalindrom(const string &s, int l, int r) {
  for (int i = 0; i < (r - l + 1) / 2; i++) {
    if (s[l + i] != s[r - i]) {
      return 0;
    }
  }
  return 1;
}
template <class type1>
type1 gcd(type1 a, type1 b) {
  type1 t;
  while (b) {
    t = b;
    b = a % b;
    a = t;
  }
  return a;
}
long long positiveMod(long long v, long long md) {
  v = v % md;
  if (v < 0) {
    v += md;
  }
  return v;
}
void solve() {
  string ss, tt;
  vector<int> s, t;
  vector<int> res;
  int n, i, isR = 0, l, r, isX = 0;
  cin >> n;
  cin >> ss >> tt;
  for (auto it : ss) {
    s.push_back(it - '0');
  }
  for (auto it : tt) {
    t.push_back(it - '0');
  }
  l = 0;
  r = n - 1;
  for (i = n - 1; i >= 0; i--) {
    if (isR == 0) {
      if ((t[i] ^ isX) == s[r]) {
        r--;
        continue;
      } else {
        if ((abs((t[i] ^ isX) - s[l]) ^ 1) == 0) {
          res.push_back(i + 1);
          isR ^= 1;
          isX ^= 1;
          l++;
        } else {
          res.push_back(1);
          res.push_back(i + 1);
          isR ^= 1;
          isX ^= 1;
          l++;
        }
      }
    } else {
      if ((t[i] ^ isX) == s[l]) {
        l++;
        continue;
      } else {
        if ((abs((t[i] ^ isX) - s[r]) ^ 1) == 0) {
          res.push_back(i + 1);
          isR ^= 1;
          isX ^= 1;
          r--;
        } else {
          res.push_back(1);
          res.push_back(i + 1);
          isR ^= 1;
          isX ^= 1;
          r--;
        }
      }
    }
  }
  cout << res.size() << " ";
  for (auto it : res) {
    cout << it << " ";
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, i;
  cin >> n;
  for (i = 0; i < n; i++) {
    solve();
  }
  return 0;
}
