#include <bits/stdc++.h>
using namespace std;
long long a[300009], b[300009], temp[300009], n;
long long merge(long long l, long long mid, long long r) {
  long long i, j, k;
  long long inv_count = 0;
  i = l;
  j = mid;
  k = l;
  while (i < mid && j <= r) {
    if (b[i] <= b[j])
      temp[k++] = b[i++];
    else {
      temp[k++] = b[j++];
      inv_count += mid - i;
    }
  }
  while (i < mid) temp[k++] = b[i++];
  while (j <= r) temp[k++] = b[j++];
  i = l;
  while (i <= r) {
    b[i] = temp[i];
    i++;
  }
  return inv_count;
}
long long inv_ct(long long l, long long r) {
  long long mid, inv_count = 0;
  if (r > l) {
    mid = (r + l) / 2;
    inv_count += inv_ct(l, mid);
    inv_count += inv_ct(mid + 1, r);
    inv_count += merge(l, mid + 1, r);
  }
  return inv_count;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  while (t--) {
    cin >> n;
    long long i = 0;
    while (i < n) cin >> a[i++];
    i = 0;
    while (i < n) {
      b[i] = a[i];
      i++;
    }
    long long sum = inv_ct(0, n - 1);
    i = 30;
    long long ans = 0, j;
    while (i > -1) {
      ans += (1ll << i);
      j = 0;
      while (j < n) {
        b[j] = a[j] ^ ans;
        j++;
      }
      j = inv_ct(0, n - 1);
      if (j < sum)
        sum = j;
      else
        ans -= (1ll << i);
      i--;
    }
    cout << sum << " " << ans << "\n";
  }
}

### Prompt

Generate 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 a[300009], b[300009], temp[300009], n;
long long merge(long long l, long long mid, long long r) {
  long long i, j, k;
  long long inv_count = 0;
  i = l;
  j = mid;
  k = l;
  while (i < mid && j <= r) {
    if (b[i] <= b[j])
      temp[k++] = b[i++];
    else {
      temp[k++] = b[j++];
      inv_count += mid - i;
    }
  }
  while (i < mid) temp[k++] = b[i++];
  while (j <= r) temp[k++] = b[j++];
  i = l;
  while (i <= r) {
    b[i] = temp[i];
    i++;
  }
  return inv_count;
}
long long inv_ct(long long l, long long r) {
  long long mid, inv_count = 0;
  if (r > l) {
    mid = (r + l) / 2;
    inv_count += inv_ct(l, mid);
    inv_count += inv_ct(mid + 1, r);
    inv_count += merge(l, mid + 1, r);
  }
  return inv_count;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  while (t--) {
    cin >> n;
    long long i = 0;
    while (i < n) cin >> a[i++];
    i = 0;
    while (i < n) {
      b[i] = a[i];
      i++;
    }
    long long sum = inv_ct(0, n - 1);
    i = 30;
    long long ans = 0, j;
    while (i > -1) {
      ans += (1ll << i);
      j = 0;
      while (j < n) {
        b[j] = a[j] ^ ans;
        j++;
      }
      j = inv_ct(0, n - 1);
      if (j < sum)
        sum = j;
      else
        ans -= (1ll << i);
      i--;
    }
    cout << sum << " " << ans << "\n";
  }
}

```