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;
const int maxn = 3e5 + 10;
char s1[maxn], s2[maxn];
int a[maxn], b[maxn];
int ans[maxn], lazy[maxn * 4];
void build(int id, int l, int r) {
  lazy[id] = 0;
  if (l == r) {
    lazy[id] = a[l];
    return;
  }
  int mid = (l + r) >> 1;
  build(id << 1, l, mid);
  build(id << 1 | 1, mid + 1, r);
}
void push_down(int id) {
  if (lazy[id] == 0) return;
  lazy[id << 1] ^= 1;
  lazy[id << 1 | 1] ^= 1;
  lazy[id] = 0;
}
void update(int id, int l, int r, int x, int y) {
  if (x <= l && y >= r) {
    lazy[id] ^= 1;
    return;
  }
  push_down(id);
  int mid = (l + r) >> 1;
  if (x <= mid) update(id << 1, l, mid, x, y);
  if (y > mid) update(id << 1 | 1, mid + 1, r, x, y);
}
int query(int id, int l, int r, int pos) {
  if (l == r) return lazy[id];
  push_down(id);
  int mid = (l + r) >> 1;
  if (pos <= mid) return query(id << 1, l, mid, pos);
  return query(id << 1 | 1, mid + 1, r, pos);
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int now = 0, n;
    scanf("%d", &n);
    scanf("%s%s", s1 + 1, s2 + 1);
    for (int i = 1; i <= n; i++) a[i] = s1[i] - '0', b[i] = s2[i] - '0';
    build(1, 1, n);
    int root = 1, ed = n;
    for (int i = n; i > 1; i--) {
      int num = query(1, 1, n, ed);
      if (num == b[i]) {
        if (root > ed)
          ed++;
        else
          ed--;
        continue;
      }
      num = query(1, 1, n, root);
      if (num == b[i]) {
        ans[++now] = 1;
        update(1, 1, n, root, root);
      }
      ans[++now] = abs(root - ed) + 1;
      if (root < ed)
        update(1, 1, n, root, ed);
      else
        update(1, 1, n, ed, root);
      swap(root, ed);
      if (root > ed)
        ed++;
      else
        ed--;
    }
    if (query(1, 1, n, root) != b[1]) ans[++now] = 1;
    printf("%d ", now);
    for (int i = 1; i <= now; i++) {
      printf("%d ", ans[i]);
    }
    printf("\n");
  }
  return 0;
}