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;
template <typename Tp>
void read(Tp &num) {
  num = 0;
  char c = getchar();
  long long f = 1;
  while (!isdigit(c)) c == '-' ? f = -1 : 1, c = getchar();
  while (isdigit(c)) num = (num << 3) + (num << 1) + (c ^ 48), c = getchar();
  num *= f;
}
template <typename Tp, typename... etc>
void read(Tp &num, etc &...more) {
  read(num);
  read(more...);
}
const long long maxn = 1e6 + 7;
long long t, n, a[maxn], b[maxn], ans1[maxn], cnt1, ans2[maxn], cnt2;
void sp(long long i, long long j) {
  long long r = a[j];
  a[j] = a[i];
  a[i] = r;
}
void fz(long long k) {
  for (register long long i = 1; i <= k / 2; ++i) sp(i, k - i + 1);
  for (register long long i = 1; i <= k; ++i) a[i] = a[i] ^ 1;
}
signed main() {
  read(t);
  while (t--) {
    read(n);
    cnt1 = 0;
    cnt2 = 0;
    char c;
    long long col1, col2;
    for (register long long i = 1; i <= n; ++i) {
      cin >> c;
      a[i] = c - '0';
    }
    for (register long long i = 1; i <= n; ++i) {
      cin >> c;
      b[i] = c - '0';
    }
    col1 = a[1];
    col2 = b[1];
    for (register long long i = 2; i <= n; ++i) {
      if (a[i] != col1) {
        ans1[++cnt1] = i - 1;
        col1 ^= 1;
      }
    }
    for (register long long i = 2; i <= n; ++i) {
      if (b[i] != col2) {
        ans2[++cnt2] = i - 1;
        col2 ^= 1;
      }
    }
    if (col1 != col2) ans1[++cnt1] = n;
    cout << cnt1 + cnt2 << endl;
    for (register long long i = 1; i <= cnt1; ++i) cout << ans1[i] << " ";
    for (register long long i = cnt2; i >= 1; --i) cout << ans2[i] << " ";
    cout << endl;
  }
  return 0;
}
