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>
const int maxn = 1e6 + 10;
using namespace std;
char b[maxn];
char a[maxn];
deque<int> q;
int main() {
  int T;
  cin >> T;
  while (T--) {
    while (q.size()) {
      q.pop_back();
    }
    int n;
    cin >> n;
    scanf("%s", b);
    scanf("%s", a);
    for (int i = 0; i < n; i++) {
      a[i] -= '0';
      b[i] -= '0';
      q.push_back(b[i]);
    }
    bool nowtail = true;
    int reversecnt = 0;
    int changcnt = 0;
    int aarr = n - 1;
    vector<int> saveans;
    while (aarr >= 0) {
      int nowval = a[aarr];
      if (a[aarr] == q.back()) {
        q.pop_back();
      } else {
        if (q.front() == nowval) {
          saveans.push_back(1);
        }
        saveans.push_back(q.size());
        q.pop_front();
        saveans.push_back(q.size());
      }
      aarr--;
    }
    int cnt = 0;
    for (int aa : saveans) {
      if (aa != 0) cnt++;
    }
    cout << cnt;
    for (int aa : saveans) {
      if (aa != 0) {
        printf(" %d", aa);
      }
    }
    printf("\n");
  }
}