#include <bits/stdc++.h>
using namespace std;
long long power(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x) % 998244353;
    y = y / 2;
    if (y != 0) x = (x * x) % 998244353;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t, n, i, j, o, e, ans = 0, num, ans1, ans2, sz, x = 0, chk;
  cin >> n;
  long long a;
  vector<long long> v;
  for (i = 0; i < n; i++) {
    cin >> a;
    v.push_back(a);
  }
  queue<vector<long long> > q;
  q.push(v);
  long long bit = 30;
  while (bit >= 0) {
    sz = q.size();
    chk = 1ll << bit;
    ans1 = ans2 = 0;
    for (i = 0; i < sz; i++) {
      v = q.front();
      q.pop();
      vector<long long> one, zero;
      o = e = 0;
      for (j = 0; j < v.size(); j++) {
        num = v[j];
        if (num & chk) {
          one.push_back(v[j]);
          o++;
          ans2 += e;
        } else {
          zero.push_back(v[j]);
          e++;
          ans1 += o;
        }
      }
      if (!one.empty()) q.push(one);
      if (!zero.empty()) q.push(zero);
    }
    ans += min(ans1, ans2);
    if (ans1 > ans2) x += chk;
    bit--;
  }
  cout << ans << " " << x << "\n";
  return 0;
}

### Prompt

Please create a solution in CPP to the following problem:

You are given an array a consisting of n non-negative integers. You have to choose a non-negative integer x and form a new array b of size n according to the following rule: for all i from 1 to n, b_i = a_i ⊕ x (⊕ denotes the operation [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR)).

An inversion in the b array is a pair of integers i and j such that 1 ≤ i < j ≤ n and b_i > b_j.

You should choose x in such a way that the number of inversions in b is minimized. If there are several options for x — output the smallest one.

Input

First line contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5) — the number of elements in a.

Second line contains n space-separated integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.

Output

Output two integers: the minimum possible number of inversions in b, and the minimum possible value of x, which achieves those number of inversions.

Examples

Input


4
0 1 3 2


Output


1 0


Input


9
10 7 9 10 7 5 5 3 5


Output


4 14


Input


3
8 10 3


Output


0 8

Note

In the first sample it is optimal to leave the array as it is by choosing x = 0.

In the second sample the selection of x = 14 results in b: [4, 9, 7, 4, 9, 11, 11, 13, 11]. It has 4 inversions:

  * i = 2, j = 3; 
  * i = 2, j = 4; 
  * i = 3, j = 4; 
  * i = 8, j = 9. 



In the third sample the selection of x = 8 results in b: [0, 2, 11]. It has no inversions.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long power(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x) % 998244353;
    y = y / 2;
    if (y != 0) x = (x * x) % 998244353;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t, n, i, j, o, e, ans = 0, num, ans1, ans2, sz, x = 0, chk;
  cin >> n;
  long long a;
  vector<long long> v;
  for (i = 0; i < n; i++) {
    cin >> a;
    v.push_back(a);
  }
  queue<vector<long long> > q;
  q.push(v);
  long long bit = 30;
  while (bit >= 0) {
    sz = q.size();
    chk = 1ll << bit;
    ans1 = ans2 = 0;
    for (i = 0; i < sz; i++) {
      v = q.front();
      q.pop();
      vector<long long> one, zero;
      o = e = 0;
      for (j = 0; j < v.size(); j++) {
        num = v[j];
        if (num & chk) {
          one.push_back(v[j]);
          o++;
          ans2 += e;
        } else {
          zero.push_back(v[j]);
          e++;
          ans1 += o;
        }
      }
      if (!one.empty()) q.push(one);
      if (!zero.empty()) q.push(zero);
    }
    ans += min(ans1, ans2);
    if (ans1 > ans2) x += chk;
    bit--;
  }
  cout << ans << " " << x << "\n";
  return 0;
}

```