#include <bits/stdc++.h>
using namespace std;
long long count_inversion_merge(vector<long long> &array, long long first,
                                long long last) {
  long long mid = (first + last) / 2;
  long long ai = first;
  long long bi = mid + 1;
  long long final[last - first + 1], finali = 0;
  long long inversion = 0, i;
  while (ai <= mid && bi <= last) {
    if (array[ai] <= array[bi]) {
      final[finali++] = array[ai++];
    } else {
      final[finali++] = array[bi++];
      inversion += mid - ai + 1;
    }
  }
  while (ai <= mid) final[finali++] = array[ai++];
  while (bi <= last) final[finali++] = array[bi++];
  for (i = 0; i < last - first + 1; i++) array[i + first] = final[i];
  return inversion;
}
long long count_inversion(vector<long long> &array, long long a, long long b) {
  long long x, y, z, mid;
  if (a >= b) return 0;
  mid = (a + b) / 2;
  x = count_inversion(array, a, mid);
  y = count_inversion(array, mid + 1, b);
  z = count_inversion_merge(array, a, b);
  return x + y + z;
}
long long n, inversions, value;
long long get_inversions(vector<long long> a) {
  return count_inversion(a, 0, (long long)(a.size()) - 1);
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  vector<long long> a(n);
  for (long long &x : a) cin >> x;
  inversions = get_inversions(a);
  for (long long bit = 30; bit >= 0; bit--) {
    long long BIT = 1 << bit;
    for (long long &x : a) x ^= BIT;
    long long invs = get_inversions(a);
    if (invs < inversions) {
      inversions = invs;
      value |= BIT;
    } else {
      for (long long &x : a) x ^= BIT;
    }
  }
  cout << inversions << ' ' << value << '\n';
}

### Prompt

Please formulate a CPP solution 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;
long long count_inversion_merge(vector<long long> &array, long long first,
                                long long last) {
  long long mid = (first + last) / 2;
  long long ai = first;
  long long bi = mid + 1;
  long long final[last - first + 1], finali = 0;
  long long inversion = 0, i;
  while (ai <= mid && bi <= last) {
    if (array[ai] <= array[bi]) {
      final[finali++] = array[ai++];
    } else {
      final[finali++] = array[bi++];
      inversion += mid - ai + 1;
    }
  }
  while (ai <= mid) final[finali++] = array[ai++];
  while (bi <= last) final[finali++] = array[bi++];
  for (i = 0; i < last - first + 1; i++) array[i + first] = final[i];
  return inversion;
}
long long count_inversion(vector<long long> &array, long long a, long long b) {
  long long x, y, z, mid;
  if (a >= b) return 0;
  mid = (a + b) / 2;
  x = count_inversion(array, a, mid);
  y = count_inversion(array, mid + 1, b);
  z = count_inversion_merge(array, a, b);
  return x + y + z;
}
long long n, inversions, value;
long long get_inversions(vector<long long> a) {
  return count_inversion(a, 0, (long long)(a.size()) - 1);
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  vector<long long> a(n);
  for (long long &x : a) cin >> x;
  inversions = get_inversions(a);
  for (long long bit = 30; bit >= 0; bit--) {
    long long BIT = 1 << bit;
    for (long long &x : a) x ^= BIT;
    long long invs = get_inversions(a);
    if (invs < inversions) {
      inversions = invs;
      value |= BIT;
    } else {
      for (long long &x : a) x ^= BIT;
    }
  }
  cout << inversions << ' ' << value << '\n';
}

```