#include <bits/stdc++.h>
using namespace std;
int n, a[300100];
long long sum[40];
void solve(vector<int> a, int bit) {
  int n = a.size();
  if (n == 0 || bit < 0) return;
  vector<int> v0, v1;
  long long num1 = 0, sum1 = 0, num0 = 0, sum0 = 0;
  for (int i = 0; i < n; i++)
    if ((a[i] >> bit) & 1)
      num1++, sum1 += num0, v1.push_back(a[i]);
    else
      num0++, sum0 += num1, v0.push_back(a[i]);
  sum[bit] += sum0 - sum1;
  solve(v0, bit - 1);
  solve(v1, bit - 1);
}
int x[300100], b[300100];
void modify(int x, int k) {
  for (; x <= n; x += (x & -x)) b[x] += k;
}
int query(int x) {
  int k = 0;
  for (; x; x ^= (x & -x)) k += b[x];
  return k;
}
int main() {
  scanf("%d", &n);
  vector<int> arr;
  arr.resize(n);
  for (int i = 0; i < n; i++) scanf("%d", &arr[i]);
  solve(arr, 30);
  int y = 0;
  for (int i = 0; i <= 30; i++)
    if (sum[i] > 0) y ^= (1 << i);
  for (int i = 0; i < n; i++) x[i] = arr[i] ^= y;
  sort(x, x + n);
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    int v = lower_bound(x, x + n, arr[i]) - x + 1;
    ans += i - query(v);
    modify(v, 1);
  }
  printf("%lld %d\n", ans, y);
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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, a[300100];
long long sum[40];
void solve(vector<int> a, int bit) {
  int n = a.size();
  if (n == 0 || bit < 0) return;
  vector<int> v0, v1;
  long long num1 = 0, sum1 = 0, num0 = 0, sum0 = 0;
  for (int i = 0; i < n; i++)
    if ((a[i] >> bit) & 1)
      num1++, sum1 += num0, v1.push_back(a[i]);
    else
      num0++, sum0 += num1, v0.push_back(a[i]);
  sum[bit] += sum0 - sum1;
  solve(v0, bit - 1);
  solve(v1, bit - 1);
}
int x[300100], b[300100];
void modify(int x, int k) {
  for (; x <= n; x += (x & -x)) b[x] += k;
}
int query(int x) {
  int k = 0;
  for (; x; x ^= (x & -x)) k += b[x];
  return k;
}
int main() {
  scanf("%d", &n);
  vector<int> arr;
  arr.resize(n);
  for (int i = 0; i < n; i++) scanf("%d", &arr[i]);
  solve(arr, 30);
  int y = 0;
  for (int i = 0; i <= 30; i++)
    if (sum[i] > 0) y ^= (1 << i);
  for (int i = 0; i < n; i++) x[i] = arr[i] ^= y;
  sort(x, x + n);
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    int v = lower_bound(x, x + n, arr[i]) - x + 1;
    ans += i - query(v);
    modify(v, 1);
  }
  printf("%lld %d\n", ans, y);
}

```