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;
const long double PI = 3.141592653589793;
const long long mod = 1e9 + 7;
int CASE = 1;
const int mxn = 1e5 + 1;
const long long infll = 1e18;
const int infi = 1e9;
bool prime(long long n) {
  if (n <= 1) return false;
  if (n == 2 or n == 3) return true;
  if (n % 2 == 0 or n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i += 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
long long __gcd(long long a, long long b) { return !b ? a : __gcd(b, a % b); }
long long power(long long a, long long b, long long MOD) {
  long long x = a, res = 1, p = b;
  while (p > 0) {
    if (p & 1) res *= x;
    x *= x;
    p >>= 1;
    x %= MOD, res %= MOD;
  }
  return res;
}
void solve() {
  int n, cnt = 1, ok = 0;
  string a, b;
  cin >> n >> a >> b;
  vector<int> ans;
  int l = 0, r = n - 1, pos = n;
  for (int i = n - 1; i >= 0; i--) {
    if (cnt) {
      if ((a[r] - '0' + ok) % 2 + '0' != b[i]) {
        if (i == 0) {
          ans.push_back(i + 1);
          continue;
        }
        if (a[l] != a[r]) ans.push_back(1), a[l] = (a[l] == '1' ? '0' : '1');
        ans.push_back(i + 1);
        l++;
        cnt ^= 1;
        ok++;
        ok %= 2;
      } else {
        r--;
      }
    } else {
      if ((a[l] - '0' + ok) % 2 + '0' != b[i]) {
        if (i == 0) {
          ans.push_back(i + 1);
          continue;
        }
        if (a[l] != a[r]) ans.push_back(1), a[r] = (a[r] == '1' ? '0' : '1');
        ans.push_back(i + 1);
        r--;
        cnt ^= 1;
        ok++;
        ok %= 2;
      } else {
        l++;
      }
    }
  }
  cout << ans.size() << ' ';
  for (int i : ans) cout << i << ' ';
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}
