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;
long long power(long long a, long long n) {
  long long p = 1;
  while (n > 0) {
    if (n % 2) {
      p = p * a;
    }
    n >>= 1;
    a *= a;
  }
  return p;
}
long long power(long long a, long long n, long long mod) {
  long long p = 1;
  while (n > 0) {
    if (n % 2) {
      p = p * a;
      p %= mod;
    }
    n >>= 1;
    a *= a;
    a %= mod;
  }
  return p % mod;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long max(long long a, long long b) {
  if (a > b) {
    return a;
  } else {
    return b;
  }
}
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
long long modInverse(long long a, long long m) { return power(a, m - 2, m); }
void out(vector<long long> arr) {
  for (auto i : arr) {
    cout << i << ' ';
  }
  cout << "\n";
}
void out(vector<pair<long long, long long>> par) {
  for (auto i : par) {
    cout << i.first << ' ';
  }
  cout << endl;
  for (auto i : par) {
    cout << i.second << ' ';
  }
  cout << endl;
}
void out(set<long long> arr) {
  for (auto i : arr) {
    cout << i << ' ';
  }
  cout << endl;
}
void inp(vector<long long> &arr) {
  for (long long i = 0; i < arr.size(); i++) {
    cin >> arr[i];
  }
}
void inp(long long &n) { cin >> n; }
void inp(long long &a, long long &b, long long &c) { cin >> a >> b >> c; }
void inp(long long &a, long long &b) { cin >> a >> b; }
void inp(long long &a, long long &b, long long &c, long long &d) {
  cin >> a >> b >> c >> d;
}
long long MOD = 998244353;
const long long N = 1000005;
void solve() {
  long long n;
  cin >> n;
  string a, b;
  cin >> a >> b;
  vector<long long> ans;
  long long len = n + 1;
  long long fl = 0;
  long long st = 0, en = n - 1;
  for (long long i = n - 1; i > 0; i--) {
    len--;
    if (fl % 2) {
      if (a[st] != b[i]) {
        st++;
        continue;
      }
      if (b[i] == '0') {
        if (a[en] == '1') {
          ans.push_back(1);
        }
        ans.push_back(len);
        fl++;
        en--;
      } else {
        if (a[en] == '0') {
          ans.push_back(1);
        }
        ans.push_back(len);
        fl++;
        en--;
      }
    } else {
      if (a[en] == b[i]) {
        en--;
        continue;
      }
      if (b[i] == '0') {
        if (a[st] == '0') {
          ans.push_back(1);
        }
        ans.push_back(len);
        fl++;
        st++;
      } else {
        if (a[st] == '1') {
          ans.push_back(1);
        }
        ans.push_back(len);
        fl++;
        st++;
      }
    }
  }
  if (fl % 2) {
    if (a[en] == b[0]) ans.push_back(1);
  } else {
    if (a[st] != b[0]) ans.push_back(1);
  }
  cout << ans.size() << "\n";
  out(ans);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}