#include <bits/stdc++.h>
using namespace std;
bool check(long long mask, long long pos) {
  return (bool)(mask & (1LL << pos));
}
long long bigmod(long long b, long long p, long long md) {
  if (p == 0) return 1;
  if (p % 2 == 1) {
    return ((b % md) * bigmod(b, p - 1, md)) % md;
  } else {
    long long y = bigmod(b, p / 2, md);
    return (y * y) % md;
  }
}
long long cnt[31][2];
void solve(vector<long long>& arr, long long b) {
  if (arr.size() == 0 || b < 0) return;
  vector<long long> v[2];
  long long c[2] = {0};
  for (long long now : arr) {
    long long bit = check(now, b);
    if (bit)
      cnt[b][1] += c[0];
    else
      cnt[b][0] += c[1];
    c[bit]++;
    v[bit].push_back(now);
  }
  solve(v[0], b - 1);
  solve(v[1], b - 1);
}
int main() {
  long long n;
  cin >> n;
  vector<long long> arr(n);
  for (long long i = 0; i < n; i++) cin >> arr[i];
  solve(arr, 30);
  long long x = 0;
  long long inv = 0;
  for (long long i = 30; i >= 0; i--) {
    inv += min(cnt[i][0], cnt[i][1]);
    if (cnt[i][0] > cnt[i][1]) {
      x += (1LL << i);
    }
  }
  cout << inv << ' ' << x << endl;
  return 0;
}

### 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;
bool check(long long mask, long long pos) {
  return (bool)(mask & (1LL << pos));
}
long long bigmod(long long b, long long p, long long md) {
  if (p == 0) return 1;
  if (p % 2 == 1) {
    return ((b % md) * bigmod(b, p - 1, md)) % md;
  } else {
    long long y = bigmod(b, p / 2, md);
    return (y * y) % md;
  }
}
long long cnt[31][2];
void solve(vector<long long>& arr, long long b) {
  if (arr.size() == 0 || b < 0) return;
  vector<long long> v[2];
  long long c[2] = {0};
  for (long long now : arr) {
    long long bit = check(now, b);
    if (bit)
      cnt[b][1] += c[0];
    else
      cnt[b][0] += c[1];
    c[bit]++;
    v[bit].push_back(now);
  }
  solve(v[0], b - 1);
  solve(v[1], b - 1);
}
int main() {
  long long n;
  cin >> n;
  vector<long long> arr(n);
  for (long long i = 0; i < n; i++) cin >> arr[i];
  solve(arr, 30);
  long long x = 0;
  long long inv = 0;
  for (long long i = 30; i >= 0; i--) {
    inv += min(cnt[i][0], cnt[i][1]);
    if (cnt[i][0] > cnt[i][1]) {
      x += (1LL << i);
    }
  }
  cout << inv << ' ' << x << endl;
  return 0;
}

```