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;
const long long mod = 1e9 + 7;
long long powmod(long long x, long long y) {
  long long t;
  for (t = 1; y; y >>= 1, x = x * x % mod)
    if (y & 1) t = t * x % mod;
  return t;
}
long long gcd(long long x, long long y) { return y ? gcd(y, x % y) : x; }
long long lcm(long long x, long long y) { return x * (y / gcd(x, y)); }
long long modd(long long a) { return (a % mod + mod) % mod; }
long long add(long long a, long long b) { return modd(modd(a) + modd(b)); }
long long mul(long long a, long long b) { return modd(modd(a) * modd(b)); }
int smask(int i, int pos) { return (i | (1 << pos)); }
int clmask(int i, int pos) { return (i & (~(1 << pos))); }
bool chmask(int i, int pos) { return (i & (1 << pos)) != 0; }
double cordist(pair<double, double> a, pair<double, double> b) {
  return sqrt(((a.first - b.first) * (a.first - b.first)) +
              ((a.second - b.second) * (a.second - b.second)));
}
long long binpow(long long a, long long b) {
  if (b == 0) return 1;
  long long res = binpow(a, b / 2);
  if (b % 2)
    return mul(mul(res, res), a);
  else
    return mul(res, res);
}
long long n, m, q, r, i, j;
string s;
char diff(char x) {
  if (x == '0') return '1';
  return '0';
}
void solve() {
  cin >> n;
  string s1, s2;
  cin >> s1 >> s2;
  int inver = 0;
  vector<int> op;
  int l = 0;
  int r = n - 1;
  for (int i = n - 1; i >= 0; i--) {
    if (inver % 2 == 0) {
      if (s1[r] == s2[i]) {
        r--;
        continue;
      } else {
        if (s1[l] == s2[i]) {
          inver++;
          op.push_back(1);
          op.push_back(i + 1);
          l++;
        } else {
          inver++;
          op.push_back(i + 1);
          l++;
        }
      }
    } else {
      if (diff(s1[l]) == s2[i]) {
        l++;
        continue;
      } else {
        if (diff(s1[r]) == s2[i]) {
          inver++;
          op.push_back(1);
          op.push_back(i + 1);
          r--;
        } else {
          inver++;
          op.push_back(i + 1);
          r--;
        }
      }
    }
  }
  cout << op.size() << " ";
  for (i = 0; i < op.size(); i++) {
    cout << op[i] << " ";
  };
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int k = 1;
  cin >> k;
  while (k--) {
    solve();
  }
  return 0;
}
