#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-8;
inline int sign(double x) { return (x > eps) - (x < -eps); }
const int maxn = 3e5 + 10, mod = 1e9 + 7, INF = 0x3f3f3f3f;
int a[maxn];
long long inv[40][2];
queue<int> q[2];
void solve(int l, int r, int p) {
  if (r - l + 1 < 2 || p < 0) return;
  int cnt[2] = {0};
  long long num0 = 0, num1 = 0;
  for (int i = l; i <= r; ++i) {
    int b = (a[i] >> p) & 1;
    if (b)
      num1 += cnt[0];
    else
      num0 += cnt[1];
    q[b].push(a[i]);
    cnt[b]++;
  }
  inv[p][1] += num1;
  inv[p][0] += num0;
  int m = l + int(q[0].size()) - 1;
  for (int i = l; i <= r; ++i) a[i] = q[i > m].front(), q[i > m].pop();
  solve(l, m, p - 1);
  solve(m + 1, r, p - 1);
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  solve(1, n, 30);
  long long ans = 0, x = 0;
  for (int i = 30; i >= 0; --i)
    ans += min(inv[i][0], inv[i][1]), x ^= int(inv[i][1] < inv[i][0]) << i;
  cout << ans << " " << x;
  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 double eps = 1e-8;
inline int sign(double x) { return (x > eps) - (x < -eps); }
const int maxn = 3e5 + 10, mod = 1e9 + 7, INF = 0x3f3f3f3f;
int a[maxn];
long long inv[40][2];
queue<int> q[2];
void solve(int l, int r, int p) {
  if (r - l + 1 < 2 || p < 0) return;
  int cnt[2] = {0};
  long long num0 = 0, num1 = 0;
  for (int i = l; i <= r; ++i) {
    int b = (a[i] >> p) & 1;
    if (b)
      num1 += cnt[0];
    else
      num0 += cnt[1];
    q[b].push(a[i]);
    cnt[b]++;
  }
  inv[p][1] += num1;
  inv[p][0] += num0;
  int m = l + int(q[0].size()) - 1;
  for (int i = l; i <= r; ++i) a[i] = q[i > m].front(), q[i > m].pop();
  solve(l, m, p - 1);
  solve(m + 1, r, p - 1);
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  solve(1, n, 30);
  long long ans = 0, x = 0;
  for (int i = 30; i >= 0; --i)
    ans += min(inv[i][0], inv[i][1]), x ^= int(inv[i][1] < inv[i][0]) << i;
  cout << ans << " " << x;
  return 0;
}

```