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;
long long read() {
  long long w = 1, q = 0;
  char ch = ' ';
  while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
  if (ch == '-') w = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') q = q * 10 + ch - '0', ch = getchar();
  return w * q;
}
const int maxn = 9 + 2e5;
const long long mod1 = 998244353;
const long long mod2 = 1000000007;
const long long mod3 = 1000000009;
int _;
long long qpow(long long m, long long n, long long mod) {
  long long temp;
  if (n == 0) return 1;
  temp = qpow(m, n >> 1, mod);
  if (n % 2 == 0)
    return temp * temp % mod;
  else
    return m * temp % mod * temp % mod;
}
int main() {
  char s1[maxn], s2[maxn];
  for (scanf("%d", &_); _; _--) {
    int ans[maxn] = {}, cnt = 0;
    int tmp = -1, n = read();
    scanf("%s%s", s1, s2);
    if (strcmp(s1, s2) == 0) {
      puts("0");
      continue;
    }
    s1[n] = 48;
    for (int i = 0; i < n; i++) {
      if (i && tmp == -1 && s1[i] == 49) ans[cnt++] = i;
      if (s1[i] == 49) tmp = 0;
      if (s1[i] == 49 && s1[i + 1] == 48) {
        ans[cnt++] = i + 1;
        tmp = -1;
      }
    }
    tmp = 0;
    for (int i = n - 1; i >= 1; i--) {
      if (s2[i] == 49) tmp++;
      if (s2[i - 1] == 48 && s2[i] == 49) {
        ans[cnt++] = i + tmp;
        ans[cnt++] = i;
        tmp = 0;
      }
    }
    if (s2[0] == 49) {
      ans[cnt++] = tmp + 1;
    }
    if (tmp && s2[0] == 48) {
      ans[cnt++] = tmp + 1;
      ans[cnt++] = 1;
    }
    printf("%d ", cnt);
    for (int i = 0; i < cnt; i++)
      printf("%d%s", ans[i], i == cnt - 1 ? "\n" : " ");
  }
  return 0;
}