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;
bool sortBySecond(const pair<long long, long long> &a,
                  const pair<long long, long long> b) {
  return a.second > b.second;
}
int ncr[1001][1001];
bool vis_ncr[1001][1001];
long long NCR(int n, int r) {
  if (n < r) return 0;
  if (vis_ncr[n][r]) return ncr[n][r];
  if (r == 0) return 1;
  if (r == 1) return n;
  if (n == 0) return 0;
  ncr[n][r] = NCR(n - 1, r) + NCR(n - 1, r - 1);
  vis_ncr[n][r] = true;
  return ncr[n][r];
}
bool isPrime[100005];
void _seive() {
  memset(isPrime, true, sizeof(isPrime));
  isPrime[1] = false;
  isPrime[2] = false;
  for (int i = 2; i * i <= 100005; i++) {
    for (int j = i * i; j <= 100005; j += i) {
      isPrime[j] = false;
    }
  }
}
long long expo(int a, int b) {
  if (b == 0)
    return 1;
  else if (b == 1)
    return a;
  else if (b % 2 == 0) {
    long long ret = expo(a, b / 2);
    ret = (ret % 1000000009 * ret % 1000000009) % 1000000009;
    return ret;
  } else {
    long long ret = expo(a, b / 2);
    ret = (ret % 1000000009 * ret % 1000000009) % 1000000009;
    ret = (ret % 1000000009 * a % 1000000009) % 1000000009;
    return ret;
  }
}
long long inverseMod(long long a, long long m) { return expo(a, m - 2) % m; }
long long fact[100005];
void factorial() {
  fact[0] = fact[1] = 1;
  for (int i = 2; i < 100005; i++)
    fact[i] = (fact[i - 1] % 1000000009 * i % 1000000009) % 1000000009;
}
long long GCD(long long a, long long b) {
  if (a == 0) return b;
  return GCD(b % a, a);
}
long long mult(long long a, long long b) {
  return (a % 1000000009 * b % 1000000009) % 1000000009;
}
long long add(long long a, long long b) {
  a += b;
  if (a >= 1000000009) return a - 1000000009;
  return a;
}
long long sub(long long a, long long b) {
  a -= b;
  if (a < 0) return a + 1000000009;
  return a;
}
int n;
string a, b;
void _run() {
  cin >> n;
  cin >> a >> b;
  vector<int> p;
  while (a != b) {
    int i = 0;
    for (; i < n; i++)
      if (a[i] != b[i]) break;
    p.push_back(i + 1);
    p.push_back(1);
    p.push_back(i + 1);
    a[i] = b[i];
  }
  cout << p.size() << " ";
  for (auto it : p) cout << it << ' ';
  cout << endl;
  ;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  int test = 1;
  cin >> test;
  while (test--) _run();
  return 0;
}
