#include <bits/stdc++.h>
using namespace std;
int n;
vector<long long> inv0(32);
vector<long long> inv1(32);
void rek(int i, vector<int> v) {
  if (i == -1 || v.size() == 0) {
    return;
  }
  vector<int> v0, v1;
  long long cnt0 = 0ll, cnt1 = 0ll;
  for (int j = 0; j < v.size(); j++) {
    int a = v[j];
    if ((a & (1 << i)) > 0) {
      cnt1++;
      inv0[i] += cnt0;
      v1.push_back(a);
    } else {
      cnt0++;
      inv1[i] += cnt1;
      v0.push_back(a);
    }
  }
  rek(i - 1, v0);
  rek(i - 1, v1);
}
int main() {
  int a;
  cin >> n;
  vector<int> v;
  for (int i = 0; i < n; i++) {
    cin >> a;
    v.push_back(a);
  }
  rek(31, v);
  long long sol1 = 0, sol2 = 0;
  for (int i = 0; i <= 31; i++) {
    sol1 += (long long)min(inv0[i], inv1[i]);
    if (inv1[i] > inv0[i]) {
      sol2 += (1 << i);
    }
  }
  cout << sol1 << " " << sol2 << endl;
}

### Prompt

Please formulate a CPP solution 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;
int n;
vector<long long> inv0(32);
vector<long long> inv1(32);
void rek(int i, vector<int> v) {
  if (i == -1 || v.size() == 0) {
    return;
  }
  vector<int> v0, v1;
  long long cnt0 = 0ll, cnt1 = 0ll;
  for (int j = 0; j < v.size(); j++) {
    int a = v[j];
    if ((a & (1 << i)) > 0) {
      cnt1++;
      inv0[i] += cnt0;
      v1.push_back(a);
    } else {
      cnt0++;
      inv1[i] += cnt1;
      v0.push_back(a);
    }
  }
  rek(i - 1, v0);
  rek(i - 1, v1);
}
int main() {
  int a;
  cin >> n;
  vector<int> v;
  for (int i = 0; i < n; i++) {
    cin >> a;
    v.push_back(a);
  }
  rek(31, v);
  long long sol1 = 0, sol2 = 0;
  for (int i = 0; i <= 31; i++) {
    sol1 += (long long)min(inv0[i], inv1[i]);
    if (inv1[i] > inv0[i]) {
      sol2 += (1 << i);
    }
  }
  cout << sol1 << " " << sol2 << endl;
}

```