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;
long long int mod = 1000000007;
vector<long long int> initializeDiffArray(vector<long long int>& A);
void update(vector<long long int>& D, long long int l, long long int r,
            long long int x);
void getArray(vector<long long int>& A, vector<long long int>& D);
long long int min(long long int a, long long int b);
long long int max(long long int a, long long int b);
long long int gcd(long long int a, long long int b);
void swap(long long int* a, long long int* b);
long long int lcm(long long int a, long long int b);
long long int modpower(long long int x, long long int y, long long int p);
long long int power(long long int x, long long int y);
long long int modulo(long long int value, long long int m);
long long int myXOR(long long int x, long long int y);
long long int diff(long long int a, long long int b);
int main() {
  {
    ios ::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
  };
  long long int rep = 1;
  cin >> rep;
  while (rep--) {
    long long int n;
    cin >> n;
    string a, b;
    cin >> a >> b;
    vector<long long int> v;
    for (int i = n - 1; i >= 0; i--) {
      if (a[i] != b[i]) {
        if (b[i] == a[0]) {
          v.push_back(1);
        }
        v.push_back(i + 1);
        a[0] = b[i];
        long long int l = 1, r = i - 1;
        while (l <= r) {
          char cl, cr;
          if (a[l] == '0')
            cl = '1';
          else
            cl = '0';
          if (a[r] == '0')
            cr = '1';
          else
            cr = '0';
          a[l] = cr;
          a[r] = cl;
          l++;
          r--;
        }
      }
    }
    cout << v.size() << '\n';
    for (auto i : v) cout << i << ' ';
    cout << '\n';
  }
  return 0;
}
long long int myXOR(long long int x, long long int y) {
  return (x | y) & (~x | ~y);
}
long long int min(long long int a, long long int b) {
  if (a > b) return b;
  return a;
}
long long int max(long long int a, long long int b) {
  if (a < b) return b;
  return a;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long int diff(long long int a, long long int b) {
  if (a > b) return a - b;
  return b - a;
}
void swap(long long int* a, long long int* b) {
  long long int t = *a;
  *a = *b;
  *b = t;
}
long long int lcm(long long int a, long long int b) {
  if (a == 0) return b;
  if (b == 0) return a;
  return (a * b) / gcd(a, b);
}
long long int modpower(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int power(long long int x, long long int y) {
  long long int temp;
  if (y == 0) return 1;
  temp = power(x, y / 2);
  if (y % 2 == 0)
    return temp * temp;
  else
    return x * temp * temp;
}
long long int modulo(long long int value, long long int m) {
  long long int mod = value % m;
  if (value < 0) {
    mod += m;
  }
  return mod;
}
vector<long long int> initializeDiffArray(vector<long long int>& A) {
  long long int n = A.size();
  vector<long long int> D(n + 1);
  D[0] = A[0], D[n] = 0;
  for (long long int i = 1; i < n; i++) D[i] = A[i] - A[i - 1];
  return D;
}
void update(vector<long long int>& D, long long int l, long long int r,
            long long int x) {
  D[l] += x;
  D[r + 1] -= x;
}
void getArray(vector<long long int>& A, vector<long long int>& D) {
  for (long long int i = 0; i < A.size(); i++) {
    if (i == 0)
      A[i] = D[i];
    else
      A[i] = D[i] + A[i - 1];
  }
}