#include <bits/stdc++.h>
using namespace std;
const int maxn = 6e6 + 5;
long long n;
long long ch[maxn][2], a[maxn], cnt;
long long size[maxn << 1], ans0[50], ans1[50];
void insert(long long x) {
  int now = 0;
  for (int i = 30; i >= 0; i--) {
    int bit = (x >> i) & 1;
    if (!ch[now][bit]) ch[now][bit] = ++cnt;
    if (bit == 0) {
      ans0[i] += size[ch[now][1]];
    } else {
      ans1[i] += size[ch[now][0]];
    }
    now = ch[now][bit];
    size[now]++;
  }
}
int main() {
  scanf("%lld", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (int i = n; i >= 1; i--) insert(a[i]);
  long long res = 0, sum = 0;
  for (int i = 0; i <= 30; i++) {
    if (ans1[i] > ans0[i]) {
      res += 1 << i;
      sum += ans0[i];
    } else {
      sum += ans1[i];
    }
  }
  cout << sum << " " << res << endl;
  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;
const int maxn = 6e6 + 5;
long long n;
long long ch[maxn][2], a[maxn], cnt;
long long size[maxn << 1], ans0[50], ans1[50];
void insert(long long x) {
  int now = 0;
  for (int i = 30; i >= 0; i--) {
    int bit = (x >> i) & 1;
    if (!ch[now][bit]) ch[now][bit] = ++cnt;
    if (bit == 0) {
      ans0[i] += size[ch[now][1]];
    } else {
      ans1[i] += size[ch[now][0]];
    }
    now = ch[now][bit];
    size[now]++;
  }
}
int main() {
  scanf("%lld", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  for (int i = n; i >= 1; i--) insert(a[i]);
  long long res = 0, sum = 0;
  for (int i = 0; i <= 30; i++) {
    if (ans1[i] > ans0[i]) {
      res += 1 << i;
      sum += ans0[i];
    } else {
      sum += ans1[i];
    }
  }
  cout << sum << " " << res << endl;
  return 0;
}

```