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 long long int MOD = 1000000007, LIM = 4294967295;
long long int n, m, a, b, c, d, q, k, tr, summ, ans, minx, maxx, l, r, mid, l1,
    r1, mid1, res, mass1[1000000], mass2[1000000], mass[1000000];
string s, trs, str, ss;
set<long long int> g, bb, all;
set<long long int> myset, myset1;
vector<long long int> myvec, myvec1, toWrite, myvec2;
vector<pair<long long int, long long int>> mm, mm1;
vector<vector<long long int>> mvv;
queue<long long int> myq;
map<long long int, long long int> mymap, mymap1, mymap2;
map<long long int, vector<long long int>> mymapvec;
multiset<long long int> mmset;
unordered_map<string, long long int> smap;
int compare(const void* a, const void* b) {
  return -((*(long long int*)a) - (*(long long int*)b));
};
long long int gcd(long long int a, long long int b) {
  while (b != 0) {
    long long int t = b;
    b = a % b;
    a = t;
  }
  return a;
}
long long int gcdex(long long int a, long long int b, long long int& x,
                    long long int& y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long int x1, y1;
  long long int d = gcdex(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
long long int NOK(long long int n1, long long int n2) {
  return n1 * n2 / gcd(n1, n2);
}
long long int bin_pow(long long int base, long long int p) {
  if (p == 1) {
    return base % MOD;
  }
  if (p == 0) {
    return 1;
  }
  if (p % 2 == 0) {
    long long int t = bin_pow(base, p / 2) % MOD;
    return t * t % MOD;
  } else {
    return bin_pow(base, p - 1) * base % MOD;
  }
}
long long int inverse_element(long long int x) { return bin_pow(x, MOD - 2); }
long long int divide(long long int a, long long int b) {
  return a * inverse_element(b) % MOD;
}
void doo(long long int i) {
  toWrite.push_back(i + 1);
  swap(mass[i], mass[i + 1]);
  swap(mass[i], mass[i + 2]);
}
long long int rec(string s, int c) { return 1; }
int main() {
  ios::sync_with_stdio(0);
  cin.tie();
  q = 1;
  cin >> q;
  while (q--) {
    cin >> n >> s >> str;
    l = 0;
    r = n - 1;
    a = 0;
    toWrite.clear();
    for (long long int i = 1; i < n; i++) {
      if ((s[r] - '0' + a) % 2 == str[n - i] - '0') {
        if (l < r) {
          r--;
        } else {
          r++;
        }
        continue;
      }
      if ((s[l] - '0' + a) % 2 != str[n - i] - '0') {
        toWrite.push_back(abs(l - r) + 1);
      } else {
        toWrite.push_back(1);
        toWrite.push_back(abs(l - r) + 1);
      }
      if (l < r) {
        l++;
      } else {
        l--;
      }
      swap(l, r);
      a++;
    }
    if ((s[l] - '0' + a) % 2 != str[0] - '0') {
      toWrite.push_back(1);
    }
    cerr << "------" << ' ' << endl;
    ;
    cout << toWrite.size() << ' ';
    for (const auto& i : toWrite) {
      cout << i << ' ';
    }
    cout << endl;
  }
  return 0;
}