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>
#pragma GCC optimize(2, 3, "Ofast")
using namespace std;
namespace fast_io {
char buf[1 << 12], *p1 = buf, *p2 = buf, sr[1 << 23], z[23], nc;
int C = -1, Z = 0, Bi = 0, Ny;
inline char gc() {
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 12, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline long long read() {
  long long x = 0;
  Ny = 1;
  while (nc = gc(), (nc < 48 || nc > 57) && nc != EOF)
    if (nc == 45) Ny = -1;
  Bi = 1;
  if (nc < 0) return nc;
  x = nc - 48;
  while (nc = gc(), 47 < nc && nc < 58 && nc != EOF)
    x = (x << 3) + (x << 1) + (nc ^ 48), Bi++;
  return x * Ny;
}
inline double gf() {
  int a = read(), y = Ny, b = (nc != '.') ? 0 : read();
  return (b ? a + (double)b / pow(10, Bi) * y : a);
}
inline int gs(char *s) {
  char c, *t = s;
  while (c = gc(), c <= 32)
    ;
  *s++ = c;
  while (c = gc(), c > 32) *s++ = c;
  return s - t;
}
inline void ot() {
  fwrite(sr, 1, C + 1, stdout);
  C = -1;
}
inline void flush() {
  if (C > 1 << 22) ot();
}
template <typename T>
inline void write(T x, char t) {
  int y = 0;
  if (x < 0) y = 1, x = -x;
  while (z[++Z] = x % 10 + 48, x /= 10)
    ;
  if (y) z[++Z] = '-';
  while (sr[++C] = z[Z], --Z)
    ;
  sr[++C] = t;
  flush();
}
inline void write(char *s) {
  int l = strlen(s);
  for (int i = 0; i < l; i++) sr[++C] = *s++;
  sr[++C] = '\n';
  flush();
}
};  // namespace fast_io
using namespace fast_io;
const int N = 2e5 + 5;
int t, n, m, o[N];
char a[N], b[N], c[N];
int main() {
  t = read();
  while (t--) {
    n = read();
    gs(a + 1);
    gs(b + 1);
    memcpy(c + 1, a + 1, 2 * n);
    m = 0;
    int l = 1, r = n, p = n, op = 1, nw = 0;
    for (int i = n; i; i--) {
      if ((a[op ? r : l] ^ nw) != b[i]) {
        if ((a[op ? l : r] ^ nw) != (b[i] ^ 1)) o[++m] = 1;
        o[++m] = i, op ^= 1, nw ^= 1;
      }
      if (op)
        r--;
      else
        l++;
    }
    write(m, ' ');
    for (int i = 1; i <= m; i++) write(o[i], " \n"[i == m]);
    memset(a + 1, 0, 2 * n);
    memset(b + 1, 0, 2 * n);
  }
  return ot(), 0;
}