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 MOD = 1e9 + 7;
const double pi = 2 * acos(0.0);
long long a1, a2, a3, b1, b2, b3, c, c1, c2, c3, d, d1, d2, d3, d4, d5, e1, e2,
    e3, f, h, k, l, m, n, m1, m2, n1, n2, o, p, p1, p2, p3, p4, q, q1, q2, q3,
    q4, r, s, t, u, v, x, y, z, sum, mx = -1e15, mn = 1e15, flg, flg1, flg2,
                                     id1, id2, id3, id4, lim, mx_id;
char str1[200001], str2[200001];
vector<long long> V;
void solve() {
  c = 0;
  flg = 0;
  id1 = 0;
  id2 = n - 1;
  V.clear();
  for (long long i = n - 1; i >= 0; i--) {
    if (flg) {
      if (str1[id1] == str2[i]) {
        if (str1[id2] != str2[i]) {
          V.push_back(1);
          c++;
        }
        V.push_back(i + 1);
        c++;
        flg = flg ^ 1;
        id2--;
      } else {
        id1++;
      }
    } else {
      if (str1[id2] != str2[i]) {
        if (str1[id1] == str2[i]) {
          V.push_back(1);
          c++;
        }
        V.push_back(i + 1);
        c++;
        flg = flg ^ 1;
        id1++;
      } else {
        id2--;
      }
    }
  }
  printf("%I64d ", c);
  for (long long i = 0; i < c; i++) printf("%I64d ", V[i]);
  printf("\n");
}
int main() {
  t = 1;
  scanf("%I64d", &t);
  while (t--) {
    scanf("%I64d", &n);
    scanf("%s%s", str1, str2);
    solve();
  }
  return 0;
}