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;
int faltu = 0;
int che = 1;
int count(long long int n) {
  if (n == 0) return faltu;
  int j = n % 10;
  if (j == 0 || j == 1)
    ;
  else
    che = 0;
  n /= 10;
  faltu++;
  return count(n);
}
int N = 1e5;
vector<bool> prime(N + 1, true);
vector<int> prim;
void pri() {
  for (int i = 2; i * i <= N; i++) {
    if (prime[i]) {
      for (int j = i * i; j <= N; j += i) {
        prime[j] = false;
      }
    }
  }
  for (int i = 2; i <= N; i++) {
    if (prime[i]) prim.push_back(i);
  }
  return;
}
long long dp[200002];
void fact() {
  dp[0] = 1;
  for (int i = 1; i < 200002; i++) {
    dp[i] = (dp[i - 1] * i) % 1000000007;
  }
}
long long moduleinverse(long long x) {
  long long res = 1;
  long long y = 1000000007 - 2;
  x = x % 1000000007;
  while (y > 0) {
    if (y & 1) res = (res * x) % 1000000007;
    y /= 2;
    x = (x * x) % 1000000007;
  }
  return res;
}
bool chec(int n, int k) {
  while (n) {
    if (n % 10 == k) return true;
    n = n / 10;
  }
  return false;
}
long long int loga(long long int a, long long int k) {
  long long int ans = 0;
  while (a >= k) {
    ans++;
    a /= k;
  }
  return ans;
}
long long int modpow(long long int x, long long int n) {
  if (n == 0) return 1;
  long long int a = modpow(x, n / 2);
  a = (a * a) % 1000000007;
  if (n % 2 == 1) a = (a * x) % 1000000007;
  return a;
}
void solve() {
  int n;
  cin >> n;
  string s1, s2;
  cin >> s1 >> s2;
  if (s1 == s2) {
    cout << 0 << "\n";
    return;
  }
  int a[n], b[n];
  for (long long i = 0; i < n; i++) {
    a[i] = int(s1[i] - '0');
    b[i] = int(s2[i] - '0');
  }
  int cnt = 0;
  vector<int> v;
  for (long long i = 0; i < n; i++) {
    if (a[n - i - 1] == b[n - i - 1])
      continue;
    else if (a[0] == b[n - i - 1]) {
      v.push_back(1);
      a[0] = abs(a[0] - 1);
      v.push_back(n - i);
      int c[n - i];
      for (long long j = 0; j < n - i; j++) {
        c[j] = abs(a[j] - 1);
      }
      for (long long j = n - i - 1; j >= 0; j--) {
        a[j] = c[n - i - 1 - j];
      }
    } else {
      v.push_back(n - i);
      int c[n - i];
      for (long long j = 0; j < n - i; j++) {
        c[j] = abs(a[j] - 1);
      }
      for (long long j = n - i - 1; j >= 0; j--) {
        a[j] = c[n - i - 1 - j];
      }
    }
  }
  cout << v.size() << "\n";
  for (long long i = 0; i < v.size(); i++) {
    cout << v[i] << " ";
  }
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int test;
  cin >> test;
  while (test--) {
    solve();
  }
}