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;
void A() {
  int t;
  cin >> t;
  while (t--) {
    int n, m;
    cin >> n >> m;
    vector<int> a(n, 0);
    vector<int> b(m, 0);
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i < m; i++) cin >> b[i];
    unordered_set<int> st;
    for (int i : b) st.insert(i);
    bool ok = false;
    int res = 0;
    for (int i = 0; i < n; i++) {
      if (st.count(a[i])) {
        ok = true;
        res = a[i];
        break;
      }
    }
    if (ok) {
      cout << "YES" << endl;
      cout << 1 << " " << res << endl;
    } else
      cout << "NO" << endl;
  }
}
void B() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> a(n, 0);
    for (int i = 0; i < n; i++) cin >> a[i];
    int flag = 0;
    int i = 0;
    while (i < n - 1) {
      if (a[i] > 1) {
        int j = i + 1;
        int cnt = 0;
        while (j < n - 1 && a[j] == 1) {
          cnt++;
          j++;
        }
        i = j;
      } else {
        flag = 1 - flag;
        i++;
      }
    }
    if (flag == 0)
      cout << "First" << endl;
    else
      cout << "Second" << endl;
  }
}
void helper(string& s, int r) {
  for (int i = 0; i <= r; i++) s[i] = (s[i] == '1' ? '0' : '1');
  int lo = 0, hi = r;
  while (lo < hi) {
    swap(s[lo], s[hi]);
    hi--;
    lo++;
  }
}
void C1() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    string a, b;
    cin >> a >> b;
    vector<int> res;
    for (int i = n - 1; i >= 0; i--) {
      if (a[i] == b[i]) continue;
      if (a[0] != b[i]) {
        helper(a, i);
        res.push_back(i);
      } else {
        helper(a, 0);
        res.push_back(0);
        helper(a, i);
        res.push_back(i);
      }
    }
    cout << res.size() << " ";
    for (auto i : res) cout << i + 1 << " ";
    cout << endl;
  }
}
void D() {}
void E() {}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  C1();
}
