#include <bits/stdc++.h>
using namespace std;
int n, stb;
long long ans, c;
long long count_now = 0, count_xr = 0;
bool y, z;
vector<int> vc;
vector<int> a;
unordered_map<int, int> mp;
void count(long long j) {
  mp.clear();
  count_now = 0, count_xr = 0;
  for (int i = 0; i < n; i++) {
    y = (a[i] >> j) & 1;
    if (y == 0) {
      if (mp.find(vc[i]) != mp.end()) count_now += mp[vc[i]];
    } else {
      mp[vc[i]]++;
    }
  }
  mp.clear();
  for (int i = 0; i < n; i++) {
    z = ((a[i] >> j) & 1) ^ 1;
    if (z == 0) {
      if (mp.find(vc[i]) != mp.end()) count_xr += mp[vc[i]];
    } else {
      mp[vc[i]]++;
    }
  }
  int f = (1 << j);
  for (int i = 0; i < n; i++) {
    y = (a[i] >> j) & 1;
    z = ((a[i] >> j) & 1) ^ 1;
    vc[i] += (count_now < count_xr ? (y * f) : (z * f));
  }
  c += min(count_now, count_xr);
  if (count_now > count_xr) {
    ans += (long long)(f);
  }
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  cin >> n;
  a.resize(n);
  vc.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    for (int j = 30; j >= 0; j--) {
      if ((a[i] >> j) & 1) {
        stb = max(stb, j);
      }
    }
  }
  for (int j = stb; j >= 0; j--) {
    count(j);
  }
  cout << c << " " << ans << "\n";
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
int n, stb;
long long ans, c;
long long count_now = 0, count_xr = 0;
bool y, z;
vector<int> vc;
vector<int> a;
unordered_map<int, int> mp;
void count(long long j) {
  mp.clear();
  count_now = 0, count_xr = 0;
  for (int i = 0; i < n; i++) {
    y = (a[i] >> j) & 1;
    if (y == 0) {
      if (mp.find(vc[i]) != mp.end()) count_now += mp[vc[i]];
    } else {
      mp[vc[i]]++;
    }
  }
  mp.clear();
  for (int i = 0; i < n; i++) {
    z = ((a[i] >> j) & 1) ^ 1;
    if (z == 0) {
      if (mp.find(vc[i]) != mp.end()) count_xr += mp[vc[i]];
    } else {
      mp[vc[i]]++;
    }
  }
  int f = (1 << j);
  for (int i = 0; i < n; i++) {
    y = (a[i] >> j) & 1;
    z = ((a[i] >> j) & 1) ^ 1;
    vc[i] += (count_now < count_xr ? (y * f) : (z * f));
  }
  c += min(count_now, count_xr);
  if (count_now > count_xr) {
    ans += (long long)(f);
  }
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  cin >> n;
  a.resize(n);
  vc.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    for (int j = 30; j >= 0; j--) {
      if ((a[i] >> j) & 1) {
        stb = max(stb, j);
      }
    }
  }
  for (int j = stb; j >= 0; j--) {
    count(j);
  }
  cout << c << " " << ans << "\n";
}

```