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;
const long long maxn = 3e5 + 7, mod = 1e9 + 7, MAXN = 1e6 + 7;
const double eps = 1e-9;
const long long INF = 1e18, inf = 1e15;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
long long n;
string a, b;
vector<long long> ans;
long long get_id(long long id, long long &shift, long long &step) {
  if (step % 2) return abs(id - shift);
  return id + shift;
}
char get_val(long long id, long long &shift, long long &step,
             long long &xor_all) {
  long long idd = get_id(id, shift, step);
  if (xor_all == 0) return a[idd];
  if (a[idd] == '0') return '1';
  return '0';
}
void change(long long N, long long &shift, long long &step,
            long long &xor_all) {
  ans.push_back(N);
  if (N == 1) {
    long long id = get_id(0, shift, step);
    if (a[id] == '0')
      a[id] = '1';
    else
      a[id] = '0';
    return;
  }
  xor_all ^= 1;
  if (step % 2 == 0)
    shift += N - 1;
  else
    shift -= (N - 1);
  ++step;
}
void solve() {
  ans.clear();
  cin >> n;
  cin >> a >> b;
  long long shift = 0;
  long long step = 0;
  long long xor_all = 0;
  for (long long i = n - 1; i >= 0; --i) {
    char val = get_val(i, shift, step, xor_all);
    char val0 = get_val(0, shift, step, xor_all);
    if (val == b[i]) continue;
    if (val0 != b[i]) {
      change(i + 1, shift, step, xor_all);
    } else {
      change(1, shift, step, xor_all);
      change(i + 1, shift, step, xor_all);
    }
  }
  cout << ans.size() << " ";
  for (auto v : ans) cout << v << " ";
  cout << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.precision(20);
  cout << fixed;
  long long t = 1;
  cin >> t;
  for (long long i = 0; i < t; ++i) {
    solve();
  }
  return 0;
}