#include <bits/stdc++.h>
using namespace std;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
const double pi = acosl(-1);
vector<vector<long long>> temp(31, vector<long long>(2));
long long ans = 0;
long long val = 0;
void rec(vector<long long> b, long long bit) {
  if (bit < 0 || b.size() == 0) return;
  long long m = b.size();
  vector<long long> one, zero;
  long long count0 = 0;
  long long count1 = 0;
  for (long long i = 0; i < (m); i++) {
    if (b[i] & (1 << bit)) {
      temp[bit][0] += count0;
      count1++;
      one.push_back(b[i]);
    } else {
      count0++;
      temp[bit][1] += count1;
      zero.push_back(b[i]);
    }
  }
  rec(one, bit - 1);
  rec(zero, bit - 1);
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < (n); i++) {
    cin >> a[i];
  }
  rec(a, 30);
  for (long long i = 30; i >= 0; i--) {
    ans += min(temp[i][0], temp[i][1]);
    if (temp[i][0] < temp[i][1]) {
      val |= (1 << i);
    }
  }
  cout << ans << ' ';
  cout << val << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long tt = 1;
  while (tt--) {
    solve();
  }
}

### 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;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
const double pi = acosl(-1);
vector<vector<long long>> temp(31, vector<long long>(2));
long long ans = 0;
long long val = 0;
void rec(vector<long long> b, long long bit) {
  if (bit < 0 || b.size() == 0) return;
  long long m = b.size();
  vector<long long> one, zero;
  long long count0 = 0;
  long long count1 = 0;
  for (long long i = 0; i < (m); i++) {
    if (b[i] & (1 << bit)) {
      temp[bit][0] += count0;
      count1++;
      one.push_back(b[i]);
    } else {
      count0++;
      temp[bit][1] += count1;
      zero.push_back(b[i]);
    }
  }
  rec(one, bit - 1);
  rec(zero, bit - 1);
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < (n); i++) {
    cin >> a[i];
  }
  rec(a, 30);
  for (long long i = 30; i >= 0; i--) {
    ans += min(temp[i][0], temp[i][1]);
    if (temp[i][0] < temp[i][1]) {
      val |= (1 << i);
    }
  }
  cout << ans << ' ';
  cout << val << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long tt = 1;
  while (tt--) {
    solve();
  }
}

```