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 gcd(long long a, long long b) { return (!a) ? b : gcd(b % a, a); }
long long lcm(long long x, long long y) { return x * (y / gcd(x, y)); }
void pArr(int arr[], int s);
void pVec(vector<int> v);
bool isPrime(int x);
int binarySearch(int arr[], int l, int r, int x);
int vBinarySearch(vector<long long> v, int l, int r, long long x);
int nChoosek(int n, int k);
int mod = 1e9 + 7;
void solve() {
  int n;
  cin >> n;
  string a, b;
  cin >> a >> b;
  int j = n - 1;
  vector<int> o;
  while (j >= 0) {
    if (a.at(0) != b.at(j)) {
      o.push_back(j + 1);
      for (int i = 0; i <= j; i++) {
        a.replace(i, 1, to_string(1 - stoi(a.substr(i, 1))));
      }
      reverse(a.begin(), a.begin() + j + 1);
    } else {
      o.push_back(1);
      o.push_back(j + 1);
      for (int i = 1; i <= j; i++) {
        a.replace(i, 1, to_string(1 - stoi(a.substr(i, 1))));
      }
      reverse(a.begin(), a.begin() + j + 1);
    }
    j--;
  }
  cout << o.size() << " ";
  for (int i = 0; i < o.size(); i++) {
    cout << o.at(i) << " ";
  }
  cout << endl;
}
int main() {
  string temp;
  int t;
  cin >> t;
  getline(cin, temp);
  while (t--) {
    solve();
  }
  return 0;
}
void pArr(int arr[], int s) {
  for (int i = 0; i < s; i++) {
    cout << arr[i] << " ";
  }
  cout << endl;
}
void pVec(vector<int> v) {
  for (int i = 0; i < v.size(); i++) {
    cout << v.at(i) << " ";
  }
  cout << endl;
}
bool isPrime(int x) {
  if (x == 1) {
    return false;
  }
  if (x == 2) {
    return true;
  }
  for (int i = 2; i <= (int)(pow(x, 0.5)); i++) {
    if (x % i == 0) {
      return false;
    }
  }
  return true;
}
int vBinarySearch(vector<long long> v, int l, int r, long long x) {
  while (l <= r) {
    int m = l + (r - l) / 2;
    if (v.at(m) == x) return m;
    if (v.at(m) < x)
      l = m + 1;
    else
      r = m - 1;
  }
  return r + 1;
}
int binarySearch(int arr[], int l, int r, int x) {
  while (l <= r) {
    int m = l + (r - l) / 2;
    if (arr[m] == x) return m;
    if (arr[m] < x)
      l = m + 1;
    else
      r = m - 1;
  }
  return r;
}
int nChoosek(int n, int k) {
  if (k > n) return 0;
  if (k * 2 > n) k = n - k;
  if (k == 0) return 1;
  int result = n;
  for (int i = 2; i <= k; ++i) {
    result *= (n - i + 1);
    result /= i;
  }
  return result;
}