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,no-stack-protector,tune=native")
#pragma gcc optimize("sse,sse2,sse3,sse4,ssse3")
#pragma gcc optimize("abm,mmx,avx,avx2,unroll-loops,fast-math,section-anchors")
using namespace std;
void fast_stream() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
vector<long long> z_function(string s) {
  long long n = s.length();
  vector<long long> z(n);
  for (long long i = 1, l = 0, r = 0; i < n; ++i) {
    if (i <= r) {
      z[i] = min(r - i + 1, z[i - l]);
    }
    while (i + z[i] < n && s[z[i]] == s[i + z[i]]) {
      ++z[i];
    }
    if (i + z[i] - 1 > r) {
      l = i;
      r = i + z[i] - 1;
    }
  }
  return z;
}
pair<long long, long long> find_max_podotrsok(vector<long long> a) {
  long long ans = a[0], ans_l = 0, ans_r = 0, sum = 0, minus_pos = -1;
  for (long long r = 0; r < a.size(); ++r) {
    sum += a[r];
    if (sum > ans) {
      ans = sum;
      ans_l = minus_pos + 1;
      ans_r = r;
    }
    if (sum == ans && ans_r == r) {
      ans_l = max(ans_l, minus_pos + 1);
    }
    if (0 >= sum) {
      sum = 0;
      minus_pos = r;
    }
  }
  return {ans_l, ans_r};
}
bool is_prost(long long n) {
  if (n == 0) {
    return true;
  }
  long long i = 2;
  while (i * i <= n) {
    if (n % i == 0) {
      return false;
    }
    ++i;
  }
  return true;
}
signed main() {
  fast_stream();
  long long tests;
  cin >> tests;
  while (tests--) {
    long long n;
    cin >> n;
    string s, t;
    cin >> s >> t;
    vector<long long> ans;
    long long idx = 0, i = 1;
    while (i < n) {
      while (i < n && s[i] == s[idx]) {
        ++i;
      }
      if (i < n) {
        ans.push_back(i);
      }
      idx = i;
    }
    if (s[n - 1] == '1') {
      ans.push_back(n);
    }
    long long now = 0;
    for (long long i = n - 1; i >= 0; --i) {
      if ('0' + now != t[i]) {
        now = 1 - now;
        ans.push_back(i + 1);
      }
    }
    cout << ans.size() << ' ';
    for (auto i : ans) {
      cout << i << ' ';
    }
    cout << '\n';
  }
}
