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;
void ritikio();
const int64_t inff = LLONG_MAX;
const int64_t minff = LLONG_MIN;
const int64_t inf = 1e18;
const int64_t pmod = 998244353;
const int64_t mod = 1e9 + 7;
const int64_t maxm = 1e9;
const int64_t maxN = 1e5 + 3;
char flip(char x) {
  char y;
  if (x == '1')
    y = '0';
  else
    y = '1';
  return y;
}
void solve() {
  int64_t n;
  cin >> n;
  string a, b;
  cin >> a >> b;
  vector<int64_t> v;
  int64_t s = 0, e = n - 1;
  int64_t cnt = 0;
  for (int64_t i = n - 1; i > -1; i--) {
    if (cnt == 0) {
      if (a[s] == b[i]) {
        v.push_back(1);
      }
      v.push_back(i + 1);
      s++;
    } else {
      if (a[e] != b[i]) {
        v.push_back(1);
      }
      v.push_back(i + 1);
      e--;
    }
    cnt = (cnt + 1) % 2;
  }
  cout << (int64_t)v.size() << " ";
  for (int64_t i = 0; i < (int64_t)v.size(); i++) cout << v[i] << " ";
  cout << endl;
}
signed main() {
  ritikio();
  int64_t T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
}
void ritikio() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}