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;
long long mod = pow(10, 9) + 7;
bool cmp(pair<long long, long long> x, pair<long long, long long> y) {
  if (x.first != y.first)
    return (x.first < y.first);
  else
    return x.second > y.second;
}
bool cmp2(long long x, long long y) { return y < x; }
void swap(long long *xp, long long *yp) {
  long long temp = *xp;
  *xp = *yp;
  *yp = temp;
}
const long long maxn = 1e5 + 10;
void print(vector<long long> v) {
  long long i, j, n = v.size();
  for (i = 0; i < n; i++) {
    cout << v[i] << " ";
  }
  cout << "\n";
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long power(long long x, long long y) {
  long long res = 1;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y / 2;
    x = (x * x);
  }
  return res;
}
int main() {
  long long int i, j, l, inc, dec, t, r;
  long long int c, d, q, e, ans = 0, temp, p, w, h, k, x, y, sum, pre, temp2, n,
                            m;
  t = 1;
  cin >> t;
  while (t--) {
    cin >> n;
    string a, b;
    cin >> a >> b;
    vector<long long> an;
    long long cnt = 0;
    long long f = 0;
    j = n - 1;
    for (i = n - 1; i >= 0; i--) {
      char ch = a[j], ch2 = a[f];
      if (cnt % 2 != 0) {
        if (ch == '1') {
          ch = '0';
        } else {
          ch = '1';
        }
        if (ch2 == '1') {
          ch2 = '0';
        } else {
          ch2 = '1';
        }
      }
      if (ch == b[i]) {
        if (cnt % 2 == 0) {
          j--;
        } else {
          j++;
        }
        continue;
      }
      if (ch2 == b[i]) an.push_back(1);
      k = abs(j - f) + 1;
      an.push_back(k);
      temp = j;
      if (cnt % 2 == 0) {
        j = f + 1;
      } else {
        j = f - 1;
      }
      f = temp;
      cnt++;
    }
    cout << an.size() << " ";
    for (i = 0; i < an.size(); i++) {
      cout << an[i] << " ";
    }
    cout << "\n";
  }
  return 0;
}
