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;
struct hash_pair {
  template <class T1, class T2>
  size_t operator()(const pair<T1, T2>& p) const {
    auto hash1 = hash<T1>{}(p.first);
    auto hash2 = hash<T2>{}(p.second);
    return hash1 ^ hash2;
  }
};
bool prime[10000001];
long long P = 1000000007 - 2;
long long factorial[10000001] = {0};
long long powerFunction(long long x, long long y) {
  long long res = 1;
  long long p = 1000000007;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
void sieveFunction(long long maxLimit) {
  memset(prime, true, sizeof(prime));
  prime[0] = prime[1] = false;
  for (long long i = 2; maxLimit >= i; i++) {
    if (prime[i]) {
      for (long long j = 2 * i; maxLimit >= j; j += i) prime[j] = false;
    }
  }
}
void factorialFunction(long long maxLimit) {
  factorial[0] = 1;
  for (long long i = 1; i <= maxLimit; i++)
    factorial[i] = (factorial[i - 1] * i) % 1000000007;
  return;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
signed main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    string a, b;
    cin >> a;
    string A = a;
    cin >> b;
    vector<long long> ans;
    long long temp = 0;
    long long nonOne = 0;
    long long one = 0;
    char a0 = a[0];
    for (long long i = n - 1; i >= 0; i--) {
      char c;
      if ((long long)ans.size() == 0) {
        c = a[i];
      } else {
        if (nonOne % 2 == 1) {
          c = a[temp - i];
        } else {
          c = a[temp + i];
        }
        if ((nonOne) % 2 == 1) {
          if (c == '0') {
            c = '1';
          } else {
            c = '0';
          }
        }
      }
      if (i == 0) {
        c = a0;
      }
      if (c != b[i]) {
        if (a0 != b[i]) {
          ans.push_back(i + 1);
          if (nonOne % 2 == 0) {
            temp += i;
          } else {
            temp -= i;
          }
          nonOne++;
        } else {
          ans.push_back(1);
          ans.push_back(i + 1);
          if (nonOne % 2 == 0) {
            temp += i;
          } else {
            temp -= i;
          }
          nonOne++;
          one++;
        }
        a0 = c;
        if (a0 == '0') {
          a0 = '1';
        } else {
          a0 = '0';
        }
      }
    }
    cout << ans.size() << ' ';
    for (long long i = 0; ans.size() > i; i++) {
      cout << ans[i] << ' ';
    }
    cout << "\n";
  }
}
