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;
template <typename T>
void read(T &x) {
  cin >> x;
}
template <typename T, typename T0>
void read(T &x, T0 &y) {
  cin >> x >> y;
}
template <typename T, typename T0, typename T1>
void read(T &x, T0 &y, T1 &z) {
  cin >> x >> y >> z;
}
template <typename T, typename T0, typename T1, typename T2>
void read(T &x, T0 &y, T1 &z, T2 &w) {
  cin >> x >> y >> z >> w;
}
template <typename T, typename T0>
void read(pair<T, T0> &p) {
  cin >> p.first >> p.second;
}
template <typename T>
void read(vector<T> &oneD, long long int n) {
  for (long long int(i) = 0; (i) < (n); ++(i)) {
    long long int x;
    read(x);
    oneD.push_back(x);
  }
}
template <typename T>
void read(T oneD[], long long int n) {
  for (long long int i = 0; i < n; i++) {
    read(oneD[i]);
  }
}
template <typename T>
void write(T &x) {
  cout << x << " ";
}
template <typename T>
void wrtline(T &x) {
  cout << x << "\n";
}
template <typename T, typename T0>
void write(T &x, T0 &y) {
  cout << x << " " << y << "\n";
}
template <typename T, typename T0, typename T1>
void write(T &x, T0 &y, T1 &z) {
  cout << x << " " << y << " " << z << "\n";
}
template <typename T, typename T0, typename T1, typename T2>
void write(T &x, T0 &y, T1 &z, T2 &w) {
  cout << x << " " << y << " " << z << " " << w << "\n";
}
template <typename T, typename T0>
void write(pair<T, T0> &p) {
  write(p.first);
  write(p.second);
  cout << "\n";
}
template <typename T>
void write(vector<T> &oneD, long long int n) {
  for (long long int i = 0; i < n; i++) {
    cout << oneD[i] << " ";
  }
  cout << "\n";
}
template <typename T>
void write(T oneD[], int n) {
  for (long long int i = 0; i < n; i++) {
    write(oneD[i]);
  }
  cout << "\n";
}
template <typename T, typename T0>
void write(map<T, T0> &mpp) {
  for (auto it : mpp) {
    write(it.first);
    cout << ": ";
    write(it.second);
    cout << "\n";
  }
  cout << "\n";
}
void write(vector<vector<long long int>> rc, long long int n, long long int m) {
  for (long long int(i) = 0; (i) < (n); ++(i)) {
    for (long long int(j) = 0; (j) < (m); ++(j)) {
      cout << rc[i][j] << " ";
    }
    cout << "\n";
  }
}
void solve() {
  long long int n;
  read(n);
  string s1, s2;
  vector<long long int> ans, tans;
  read(s1, s2);
  char ch = s1[0];
  for (long long int(i) = 0; (i) < (n); ++(i)) {
    if (ch != s1[i]) {
      ch = s1[i];
      ans.push_back(i);
    }
  }
  if (s1[n - 1] == '1') ans.push_back(n);
  ch = s2[0];
  for (long long int(i) = 0; (i) < (n); ++(i)) {
    if (ch != s2[i]) {
      ch = s2[i];
      tans.push_back(i);
    }
  }
  if (s2[n - 1] == '1') tans.push_back(n);
  reverse((tans).begin(), (tans).end());
  for (long long int(i) = 0; (i) < (tans.size()); ++(i)) ans.push_back(tans[i]);
  cout << ans.size() << " ";
  write(ans, ans.size());
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  read(t);
  while (t--) solve();
  return 0;
}