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>
using namespace std;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, 1, 0, -1};
const int dxo[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
const int dyo[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
const int inf = 1000000000;
const long long inf_ll = (long long)1 << 60;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = 10 * x + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline long long readll() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = 10 * x + ch - '0';
    ch = getchar();
  }
  return x * f;
}
long long gcd(long long x, long long y) { return y ? gcd(y, x % y) : x; }
long long power(long long x, long long y) {
  long long t = 1;
  for (; y; y >>= 1, x = x * x % (1000000007))
    if (y & 1) t = t * x % (1000000007);
  return t;
}
char a[(500000 + 5)], b[(500000 + 5)];
void solve() {
  vector<int> ans;
  int n = read();
  scanf("%s", a);
  scanf("%s", b);
  int left = a[0] - '0';
  int reverse_count = 0;
  int reverse_sum = 0;
  for (int i = (n - 1); i >= (1); i--) {
    int right;
    if (reverse_count % 2 == 0) {
      right = a[reverse_sum + i] - '0';
    } else {
      right = 1 - (a[reverse_sum - i] - '0');
    }
    if (right != b[i] - '0') {
      if (left == b[i] - '0') {
        ans.push_back(1);
      }
      ans.push_back(i + 1);
      if (reverse_count % 2 == 0)
        reverse_sum += i;
      else
        reverse_sum -= i;
      reverse_count++;
      left = 1 - right;
    }
  }
  if (left != b[0] - '0') ans.push_back(1);
  cout << ans.size();
  for (int i = (0); i <= ((int)ans.size() - 1); i++) cout << ' ' << ans[i];
  cout << endl;
}
int main() {
  int T = read();
  while (T--) {
    solve();
  }
  return 0;
}
