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;
using ll = long long int;
const ll mod = 1000000007;
const ll M = 1000000000000000000ll;
inline ll add(ll a, ll b) { return ((a % mod + b % mod) % mod); }
inline ll sub(ll a, ll b) { return ((a % mod - b % mod + mod) % mod); }
inline ll mult(ll a, ll b) { return (((a % mod) * (b % mod)) % mod); }
inline ll power(ll n, ll x) {
  ll p = 1;
  while (x != 0) {
    if (x % 2 == 0) {
      n = mult(n, n);
      x = x / 2;
    }
    p = mult(p, n);
    x--;
  }
  return p;
}
inline ll recurmult(ll a, ll b) {
  ll r = 0;
  while (b) {
    if (b & 1) r = add(r, a);
    a = mult(2, a);
    b >>= 1;
  }
  return r % mod;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  ll t;
  cin >> t;
  while (t--) {
    ll n;
    cin >> n;
    string s1, s2, orig;
    cin >> s1 >> s2;
    orig = s1;
    vector<ll> ans;
    for (ll i = n - 1; i >= 0; i--) {
      if (s1[i] != s2[i]) {
        if (s1[0] != s2[i]) {
          for (ll j = 0; j <= i; j++) {
            if (s1[j] == '0')
              s1[j] = '1';
            else
              s1[j] = '0';
          }
          reverse(s1.begin(), s1.begin() + i + 1);
          ans.push_back(i + 1);
        } else {
          if (s1[0] == '1')
            s1[0] = '0';
          else
            s1[0] = '1';
          ans.push_back(1);
          for (ll j = 0; j <= i; j++) {
            if (s1[j] == '0')
              s1[j] = '1';
            else
              s1[j] = '0';
          }
          reverse(s1.begin(), s1.begin() + i + 1);
          ans.push_back(i + 1);
        }
      }
    }
    cout << ans.size() << " ";
    for (auto it : ans) cout << it << " ";
    cout << "\n";
  }
}
