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;
ostream& operator<<(ostream& a, const vector<long long>& b) {
  for (auto k : b) cout << k << " ";
  return a;
}
istream& operator>>(istream& a, pair<long long, long long>& b) {
  cin >> b.first >> b.second;
  return a;
}
ostream& operator<<(ostream& a, const pair<long long, long long>& b) {
  cout << "{" << b.first << ", " << b.second << "}";
  return a;
}
template <typename T1, typename T2>
inline bool chkmin(T1& x, const T2& y) {
  if (x > y) {
    x = y;
    return 1;
  }
  return 0;
}
template <typename T1, typename T2>
inline bool chkmax(T1& x, const T2& y) {
  if (x < y) {
    x = y;
    return 1;
  }
  return 0;
}
const long long INF = 1e18;
const long double PI = acos(-1);
mt19937 gen(chrono::high_resolution_clock::now().time_since_epoch().count());
void w(long long x, vector<long long>& a) {
  reverse(a.begin(), a.begin() + x);
  for (long long i = 0; i < x; i++) {
    a[i] ^= 1;
  }
}
void solve() {
  long long n;
  cin >> n;
  string s1, s2;
  cin >> s1 >> s2;
  vector<long long> a(n), b(n);
  for (long long i = 0; i < n; i++) {
    a[i] = s1[i] - '0';
    b[i] = s2[i] - '0';
  }
  vector<long long> ans;
  long long l = 0, r = n - 1;
  long long x = 0;
  for (long long i = n - 1; i > 0; i--) {
    if ((n - 1 - i) % 2 == 0) {
      if (b[i] == (a[l] ^ x)) {
        ans.push_back(1);
        ans.push_back(i + 1);
        l++;
      } else {
        ans.push_back(i + 1);
        l++;
      }
    } else {
      if (b[i] == (a[r] ^ x)) {
        ans.push_back(1);
        ans.push_back(i + 1);
        r--;
      } else {
        ans.push_back(i + 1);
        r--;
      }
    }
    x ^= 1;
  }
  57;
  57;
  if (b[0] != (a[l] ^ x)) {
    57;
    ans.push_back(1);
  }
  57;
  cout << ans.size() << " " << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long q = 1;
  cin >> q;
  while (q--) {
    solve();
  }
}