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 comment(linker, "/STACK:102400000,102400000")
using namespace std;
#pragma optimize("-O3")
inline long long read() {
  long long x = 0;
  int f = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  return x = f ? -x : x;
}
inline void write(long long x, bool f) {
  if (x == 0) {
    putchar('0');
    if (f) putchar('\n');
    return;
  }
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  static char s[23];
  int l = 0;
  while (x != 0) s[l++] = x % 10 + 48, x /= 10;
  while (l) putchar(s[--l]);
  if (f) putchar('\n');
}
int lowbit(int x) { return x & (-x); }
template <class T>
T big(const T &a1, const T &a2) {
  return a1 > a2 ? a1 : a2;
}
template <class T>
T sml(const T &a1, const T &a2) {
  return a1 < a2 ? a1 : a2;
}
template <typename T, typename... R>
T big(const T &f, const R &...r) {
  return big(f, big(r...));
}
template <typename T, typename... R>
T sml(const T &f, const R &...r) {
  return sml(f, sml(r...));
}
void debug_out() { cout << '\n'; }
template <typename T, typename... R>
void debug_out(const T &f, const R &...r) {
  cout << f << " ";
  debug_out(r...);
}
const char ptout[] = "%d";
template <typename T>
void print(const T &f) {
  printf(ptout, f);
  putchar('\n');
}
template <typename T, typename... R>
void print(const T &f, const R &...r) {
  printf(ptout, f);
  putchar(' ');
  print(r...);
}
const int HMOD[] = {1000000009, 1004535809};
const long long BASE[] = {1572872831, 1971536491};
const long long INFLL = 0x3f3f3f3f3f3f3f3fLL;
const int INF = 0x3f3f3f3f;
const int mod = 998244353;
const int MOD = 1e9 + 7;
const int MXN = 2e5 + 5;
const int MXE = 2e6 + 6;
int n, m, k;
char s[MXN], t[MXN];
int is[MXN];
int main() {
  int tim;
  scanf("%d", &tim);
  vector<int> vs;
  while (tim--) {
    n = read();
    vs.clear();
    scanf("%s%s", s, t);
    int f = 1;
    for (int i = 0; i < n; ++i)
      if (s[i] != t[i]) f = 0;
    if (f) {
      printf("0\n");
      continue;
    }
    int cnt = 0, fir = 0;
    for (int i = 1; i < n; ++i) {
      if (s[i] != s[i - 1]) {
        ++cnt;
        fir = i;
        is[i] = 1;
        vs.emplace_back(i - 1);
      }
    }
    if (s[n - 1] == '1') vs.emplace_back(n - 1);
    for (int i = n - 1; i >= 0; --i) {
      if (i == n - 1 && t[i] == '1') vs.emplace_back(i);
      if (i && t[i] != t[i - 1]) vs.emplace_back(i - 1);
    }
    printf("%d ", vs.size());
    for (int x : vs) printf("%d ", x + 1);
    printf("\n");
  }
  return 0;
}