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;
double tick() {
  static clock_t oldt, newt = clock();
  double diff = 1.0 * (newt - oldt) / CLOCKS_PER_SEC;
  oldt = newt;
  return diff;
}
inline long long int read() {
  long long int f = 1, x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long int fastExpo(long long int a, long long int n) {
  long long int res = 1;
  while (n) {
    if (n % 2) res = res * a;
    n /= 2;
    a = a * a;
  }
  return res;
}
int dx[] = {-1, 0, 1, 0};
int dy[] = {0, 1, 0, -1};
vector<int> res;
int main() {
  int t = read();
  while (t--) {
    int n = read();
    string a, b;
    cin >> a >> b;
    int top = a[0] - '0';
    for (int i = 0; i < n; i++) {
      if ((a[i] - '0') != top) res.push_back(i), top = 1 - top;
    }
    for (int i = n - 1; i >= 0; i--) {
      if ((b[i] - '0') != top) res.push_back(i + 1), top = 1 - top;
    }
    cout << res.size() << " ";
    for (int j : res) cout << j << " ";
    cout << endl;
    res.clear();
  }
  cerr << "execution time is: " << tick() << "\n";
}