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;
inline long long read() {
  register long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + ch - '0';
    ch = getchar();
  }
  return (f == 1) ? x : -x;
}
int REVI, REVW, n, a[1010101], b[1010101], A[1010101];
char s1[1011101], s2[1010101];
vector<int> as;
int ONE;
inline void doit(int i) { as.push_back(i); }
int REV, REVV;
signed main() {
  int T = read();
  while (T--) {
    n = read();
    scanf("%s", s1 + 1);
    scanf("%s", s2 + 1);
    for (int i = 1; i <= n; i++) {
      a[i] = s1[i] - '0', b[i] = s2[i] - '0';
      A[i] = a[i];
    }
    as.clear();
    REVW = 0;
    REV = 0;
    REVV = 1;
    ONE = a[1];
    for (int i = n; i >= 2; i--) {
      if ((a[REVV * i - REV * REVV] ^ REVW) != b[i]) {
        if (ONE != b[i]) {
          ONE = a[REVV * i - REV * REVV] ^ REVW ^ 1;
          doit(i);
          REVW ^= 1;
          REV = i + 1 - REV;
          REVV = -REVV;
        } else {
          doit(1);
          ONE = a[REVV * i - REV * REVV] ^ REVW ^ 1;
          doit(i);
          REVW ^= 1;
          REV = i + 1 - REV;
          REVV = -REVV;
        }
      }
    }
    if (ONE != b[1]) {
      doit(1);
    }
    printf("%d ", (int)as.size());
    for (auto x : as) {
      printf("%d ", x);
    }
    puts("");
  }
  return 0;
}
