#include <bits/stdc++.h>
using namespace std;
const int N = 1000007, inf = 0x3f3f3f3f;
struct bit {
  vector<int> seg;
  bit() {}
  bit(int n) { seg.assign(n + 5, 0); }
  int query(int p) {
    int ans = 0;
    for (p++; p > 0; p -= p & -p) {
      ans += seg[p];
    }
    return ans;
  }
  int query(int l, int r) { return query(r) - query(l - 1); }
  void update(int p, int val) {
    for (p++; p < seg.size(); p += p & -p) seg[p] += val;
  }
  void clear() { fill(seg.begin(), seg.end(), 0); }
};
unordered_map<int, array<vector<int>, 2> > pref;
int a[N];
long long f[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> A;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    A.push_back(a[i]);
  }
  sort(A.begin(), A.end());
  A.erase(unique(A.begin(), A.end()), A.end());
  long long inv = 0;
  bit myseg(n);
  for (int i = n - 1; i >= 0; --i) {
    int x = lower_bound(A.begin(), A.end(), a[i]) - A.begin();
    inv += myseg.query(x - 1);
    myseg.update(x, +1);
  }
  bit seg(n);
  for (int b = 0; b < 30; ++b) {
    for (int i = 0; i < n; ++i) {
      pref[a[i] >> (b + 1)][(a[i] >> b) & 1].push_back(i);
    }
    for (auto& [x, y] : pref) {
      for (int p : y[0]) seg.update(p, +1);
      for (int p : y[1]) {
        f[b] += seg.query(p) - seg.query(p + 1, n);
      }
      for (int p : y[0]) seg.update(p, -1);
    }
    pref.clear();
  }
  long long ans = 0, men = 0;
  for (int i = 30; i >= 0; --i) {
    ans <<= 1;
    if (f[i] < 0) {
      men += f[i];
      ans ^= 1;
    }
  }
  cout << inv + men << ' ' << ans << endl;
}

### 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;
const int N = 1000007, inf = 0x3f3f3f3f;
struct bit {
  vector<int> seg;
  bit() {}
  bit(int n) { seg.assign(n + 5, 0); }
  int query(int p) {
    int ans = 0;
    for (p++; p > 0; p -= p & -p) {
      ans += seg[p];
    }
    return ans;
  }
  int query(int l, int r) { return query(r) - query(l - 1); }
  void update(int p, int val) {
    for (p++; p < seg.size(); p += p & -p) seg[p] += val;
  }
  void clear() { fill(seg.begin(), seg.end(), 0); }
};
unordered_map<int, array<vector<int>, 2> > pref;
int a[N];
long long f[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> A;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    A.push_back(a[i]);
  }
  sort(A.begin(), A.end());
  A.erase(unique(A.begin(), A.end()), A.end());
  long long inv = 0;
  bit myseg(n);
  for (int i = n - 1; i >= 0; --i) {
    int x = lower_bound(A.begin(), A.end(), a[i]) - A.begin();
    inv += myseg.query(x - 1);
    myseg.update(x, +1);
  }
  bit seg(n);
  for (int b = 0; b < 30; ++b) {
    for (int i = 0; i < n; ++i) {
      pref[a[i] >> (b + 1)][(a[i] >> b) & 1].push_back(i);
    }
    for (auto& [x, y] : pref) {
      for (int p : y[0]) seg.update(p, +1);
      for (int p : y[1]) {
        f[b] += seg.query(p) - seg.query(p + 1, n);
      }
      for (int p : y[0]) seg.update(p, -1);
    }
    pref.clear();
  }
  long long ans = 0, men = 0;
  for (int i = 30; i >= 0; --i) {
    ans <<= 1;
    if (f[i] < 0) {
      men += f[i];
      ans ^= 1;
    }
  }
  cout << inv + men << ' ' << ans << endl;
}

```