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;
const long long int MOD = 1000000000 + 7;
const long long int MOD1 = 998244353;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int strTo(string s) {
  stringstream tmp(s);
  long long int x = 0;
  tmp >> x;
  return x;
}
string NoTo(long long int n) {
  ostringstream str1;
  str1 << n;
  string ans = str1.str();
  return ans;
}
vector<bool> sieve(1000000, true);
void Sieve() {
  sieve[0] = false;
  sieve[1] = false;
  for (long long int i = 2; i * i <= 1000000; i++) {
    if (sieve[i] == true) {
      for (long long int j = i * i; j < 1000000; j += i) sieve[j] = false;
    }
  }
}
long long int lcm(long long int a, long long int b) {
  return (a * b) / gcd(a, b);
}
long long int arrlcm(long long int arr[], long long int n) {
  long long int ans = arr[0];
  for (long long int i = 1; i < n; i++)
    ans = (((arr[i] * ans)) / (gcd(arr[i], ans)));
  return ans;
}
long long int arrgcd(long long int arr[], long long int n) {
  long long int result = arr[0];
  for (long long int i = 1; i < n; i++) {
    result = gcd(arr[i], result);
    if (result == 1) {
      return 1;
    }
  }
  return result;
}
long long int nCr(long long int n, long long int r) {
  long long int ans = 1;
  for (long long int i = (long long int)0; i < (long long int)r; i++) {
    ans *= n - i;
    ans /= i + 1;
  }
  return ans;
}
long long int fact(long long int n) {
  if (n == 0) return 1;
  return n * fact(n - 1);
}
vector<long long int> v;
void Divisors(long long int n) {
  for (long long int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i)
        v.push_back(i);
      else {
        v.push_back(i);
        v.push_back(n / i);
      }
    }
  }
}
long long int no_of_divisors(long long int n) {
  long long int c = 0;
  for (long long int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i)
        c = c + 1;
      else
        c = c + 2;
    }
  }
  return c;
}
long long int pwr(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = res * a;
    a *= a;
    b >>= 1;
  }
  return res;
}
long long int pwrm(long long int a, long long int b, long long int m) {
  a %= m;
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
long long int digits(long long int n) {
  long long int x = floor(log10(n)) + 1;
  return x;
}
vector<long long int> prf;
void primeFactors(long long int n) {
  if (n % 2 == 0) prf.push_back(2);
  while (n % 2 == 0) {
    n = n / 2;
  }
  for (long long int i = 3; i <= sqrt(n); i = i + 2) {
    if (n % i == 0) prf.push_back(i);
    while (n % i == 0) {
      n = n / i;
    }
  }
  if (n > 1) prf.push_back(n);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int T;
  cin >> T;
  while (T--) {
    long long int n;
    cin >> n;
    string a, b;
    cin >> a >> b;
    if (a == b)
      cout << "0\n";
    else {
      vector<long long int> v;
      for (long long int i = (long long int)n - 1; i >= (long long int)0; i--) {
        if (a[i] != b[i]) {
          if (b[i] == a[0]) {
            v.push_back(1);
            v.push_back(i + 1);
            if (a[0] == '1')
              a[0] = '0';
            else
              a[0] = '1';
            for (long long int j = (long long int)0; j < (long long int)i + 1;
                 j++) {
              if (a[j] == '1')
                a[j] = '0';
              else
                a[j] = '1';
            }
            for (long long int j = (long long int)0;
                 j < (long long int)i / 2 + 1; j++)
              swap(a[j], a[i - j]);
          } else {
            v.push_back(i + 1);
            for (long long int j = (long long int)0; j < (long long int)i + 1;
                 j++) {
              if (a[j] == '1')
                a[j] = '0';
              else
                a[j] = '1';
            }
            for (long long int j = (long long int)0;
                 j < (long long int)i / 2 + 1; j++)
              swap(a[j], a[i - j]);
          }
        }
      }
      cout << v.size() << " ";
      for (auto it : v) cout << it << " ";
      cout << "\n";
    }
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
  return 0;
}