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>
#pragma GCC optimize("O3")
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
template <class T>
using V = vector<T>;
const ll inf = (1e18);
const ll mod = 1000000007;
ll GCD(ll a, ll b) { return b ? GCD(b, a % b) : a; }
ll LCM(ll c, ll d) { return c / GCD(c, d) * d; }
struct __INIT {
  __INIT() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);
  }
} __init;
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
void pitsu() {
  int n;
  cin >> n;
  string s, t;
  cin >> s >> t;
  V<int> a(n), b(n);
  for (int i = 0; i < n; i++) {
    a[i] = s[i] - '0';
    b[i] = t[i] - '0';
  }
  int m = n - 1;
  V<int> ans;
  bool rev = true;
  int l = 0, r = n - 1;
  int cnt = 0;
  while (m >= 0) {
    if (rev) {
      if ((a[r] ^ ((cnt) % 2) == b[m])) {
        r--;
        m--;
        continue;
      }
    } else {
      if ((a[l] ^ ((cnt) % 2)) == b[m]) {
        l++;
        m--;
        continue;
      }
    }
    if (a[l] != a[r]) {
      ans.push_back(1);
    }
    ans.push_back(m + 1);
    cnt++;
    if (rev)
      l++;
    else
      r--;
    rev ^= 1;
    m--;
  }
  cout << ans.size() << " ";
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i] << " ";
  }
  cout << "\n";
}
int main() {
  int t;
  cin >> t;
  while (t--) pitsu();
}
