#include <bits/stdc++.h>
using namespace std;
const long long int mod = 1e9 + 7;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int powm(long long int a, long long int b) {
  long long int res = 1;
  while (b) {
    if (b & 1) res = (res * 1LL * a) % mod;
    b >>= 1LL;
    a = (a * 1LL * a) % mod;
  }
  return res;
}
const int N = 1e6 + 5;
const int M = 2005;
long long int cnt[40][2];
void go(vector<long long int> a, int b) {
  if (a.size() == 0 || b < 0) return;
  vector<long long int> on, off;
  int one = 0, zero = 0;
  for (auto v : a) {
    if (v & (1 << b)) {
      cnt[b][1] += zero;
      one++;
      on.push_back(v);
    } else {
      cnt[b][0] += one;
      zero++;
      off.push_back(v);
    }
  }
  go(on, b - 1);
  go(off, b - 1);
}
void solve() {
  int n;
  cin >> n;
  vector<long long int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  go(a, 30);
  long long int ans = 0, inv = 0;
  for (long long int i = 30; i >= 0; i--) {
    inv += min(cnt[i][0], cnt[i][1]);
    if (cnt[i][0] > cnt[i][1]) {
      ans += (1LL << i);
    }
  }
  cout << inv << " " << ans << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int tc = 1;
  int _ = 0;
  while (_++ < tc) {
    solve();
  }
  return 0;
}

### 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 long long int mod = 1e9 + 7;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int powm(long long int a, long long int b) {
  long long int res = 1;
  while (b) {
    if (b & 1) res = (res * 1LL * a) % mod;
    b >>= 1LL;
    a = (a * 1LL * a) % mod;
  }
  return res;
}
const int N = 1e6 + 5;
const int M = 2005;
long long int cnt[40][2];
void go(vector<long long int> a, int b) {
  if (a.size() == 0 || b < 0) return;
  vector<long long int> on, off;
  int one = 0, zero = 0;
  for (auto v : a) {
    if (v & (1 << b)) {
      cnt[b][1] += zero;
      one++;
      on.push_back(v);
    } else {
      cnt[b][0] += one;
      zero++;
      off.push_back(v);
    }
  }
  go(on, b - 1);
  go(off, b - 1);
}
void solve() {
  int n;
  cin >> n;
  vector<long long int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  go(a, 30);
  long long int ans = 0, inv = 0;
  for (long long int i = 30; i >= 0; i--) {
    inv += min(cnt[i][0], cnt[i][1]);
    if (cnt[i][0] > cnt[i][1]) {
      ans += (1LL << i);
    }
  }
  cout << inv << " " << ans << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int tc = 1;
  int _ = 0;
  while (_++ < tc) {
    solve();
  }
  return 0;
}

```