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")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
using namespace std;
long long MOD = 998244353;
double eps = 1e-12;
long long myceil(long long a, long long b) {
  if (a % b == 0)
    return a / b;
  else
    return a / b + 1;
}
bool isPerSquare(long double a) {
  if (a < 0) return false;
  long long sr = sqrt(a);
  return (sr * sr == a);
}
long long arr1[200002];
long long arr2[200002];
long long arr2D[1001][1001];
vector<long long> vec;
void changestr(string &s, long long k) {
  for (long long i = 0; i < k + 1; i++) {
    if (s[i] == '0')
      s[i] = '1';
    else
      s[i] = '0';
  }
  reverse(&s[0], &s[k + 1]);
}
void solve() {
  long long n, m, a, b, c, d;
  string s1, s2;
  cin >> n;
  cin >> s1 >> s2;
  vector<long long> vec;
  for (long long i = n - 1; i >= 0; i--) {
    if (s1[i] == s2[i]) continue;
    if (s1[0] != s2[i]) {
      vec.push_back(i);
      changestr(s1, i);
      continue;
    }
    vec.push_back(0);
    vec.push_back(i);
    changestr(s1, 0);
    changestr(s1, i);
  }
  cout << vec.size() << " ";
  for (auto itr = vec.begin(); itr != vec.end(); ++itr) cout << *itr + 1 << " ";
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  cin >> t;
  for (int it = 1; it <= t; it++) {
    solve();
  }
  return 0;
}
