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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
using ull = unsigned long long;
using ll = long long;
using ld = long double;
const long long brad = 1e9 + 7;
const long long maxn = 2e5;
const string alphabet = "abcdefghijklmnopqrstuvwxyz";
struct Group {
  long long a, b, c;
};
void foo(vector<long long> a, vector<long long> b, long long n) {
  vector<long long> r;
  long long lo, hi, last;
  lo = 0;
  hi = last = n - 1;
  while (1) {
    while (last >= 0) {
      if (a[hi] != b[last]) break;
      hi--;
      last--;
    }
    if (last < 0) break;
    if (a[lo] == b[last]) r.push_back(1);
    lo++;
    r.push_back(last + 1);
    last--;
    while (last >= 0) {
      if (a[lo] ^ 1 != b[last]) break;
      lo++;
      last--;
    }
    if (last < 0) break;
    if (a[hi] ^ 1 == b[last]) r.push_back(1);
    hi--;
    r.push_back(last + 1);
    last--;
  }
  cout << r.size() << " ";
  for (auto i : r) cout << i << " ";
  cout << "\n";
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t, n, u, v, r, x, y;
  bool f;
  cin >> t;
  while (t--) {
    string s, p;
    vector<long long> a, b;
    cin >> n >> s >> p;
    for (long long i = 0; i < n; i++) {
      a.push_back(s[i] - '0');
      b.push_back(p[i] - '0');
    }
    foo(a, b, n);
  }
  return 0;
}