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() {
  long long x = 0;
  char ch = getchar();
  bool d = 1;
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') d = 0;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  return d ? x : -x;
}
inline void write(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x >= 10) write(x / 10);
  putchar('0' + x % 10);
}
inline void writeln(long long x) {
  write(x);
  puts("");
}
inline void writep(long long x) {
  write(x);
  putchar(' ');
}
inline unsigned long long rnd() {
  return ((unsigned long long)rand() << 30 ^ rand()) << 4 | rand() % 4;
}
const int N = 1e5 + 5;
char s[N], t[N];
int a[N], b[N], pos[N];
vector<long long> v;
int main() {
  int T = read();
  while (T--) {
    int n = read();
    scanf("%s%s", s + 1, t + 1);
    v.clear();
    for (int i = (int)(1); i <= (int)(n); i++)
      a[i] = s[i] - '0', b[i] = t[i] - '0';
    bool d = 0;
    int l = 1, r = n;
    for (int i = (int)(n); i >= (int)(1); i--) {
      if (!d) {
        if ((a[r] ^ d) == b[i]) {
          r--;
          continue;
        }
        if ((a[l] ^ d) == b[i]) a[l] ^= 1, v.push_back(1);
        v.push_back(i);
        l++;
        d ^= 1;
      } else {
        if ((a[l] ^ d) == b[i]) {
          l++;
          continue;
        }
        if ((a[r] ^ d) == b[i]) a[r] ^= 1, v.push_back(1);
        v.push_back(i);
        r--;
        d ^= 1;
      }
    }
    cout << (int)v.size() << " ";
    for (auto x : v) writep(x);
    puts("");
  }
}
