#include <bits/stdc++.h>
const int maxn = 4e5 + 3;
using ll = long long;
int n, a[maxn], pm, x;
ll calc(int k, int l, int r) {
  ll p1c = 0, ans = 0;
  for (int i = l; i < r; ++i) {
    if (a[i] & (1 << k))
      p1c += 1;
    else
      ans += p1c;
  }
  return ans;
}
void update(int k, int l, int r, ll &s0, ll &s1) {
  s0 += calc(k, l, r);
  for (int i = l; i < r; ++i) a[i] ^= (1 << k);
  s1 += calc(k, l, r);
  for (int i = l; i < r; ++i) a[i] ^= (1 << k);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", a + i);
  }
  ll ans = 0;
  for (int k = 30; k >= 0; --k) {
    ll s0 = 0, s1 = 0;
    for (int i = 0, j; i < n; i = j) {
      for (j = i; j < n && (a[j] & pm) == (a[i] & pm); ++j)
        ;
      update(k, i, j, s0, s1);
    }
    if (s1 < s0) {
      x |= 1 << k;
      ans += s1;
      for (int i = 0; i < n; ++i) {
        a[i] ^= (1 << k);
      }
    } else {
      ans += s0;
    }
    pm |= 1 << k;
    std::stable_sort(a, a + n,
                     [](int i, int j) { return (i & pm) < (j & pm); });
  }
  printf("%lld %d\n", ans, x);
  return 0;
}

### Prompt

Your task is to create 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>
const int maxn = 4e5 + 3;
using ll = long long;
int n, a[maxn], pm, x;
ll calc(int k, int l, int r) {
  ll p1c = 0, ans = 0;
  for (int i = l; i < r; ++i) {
    if (a[i] & (1 << k))
      p1c += 1;
    else
      ans += p1c;
  }
  return ans;
}
void update(int k, int l, int r, ll &s0, ll &s1) {
  s0 += calc(k, l, r);
  for (int i = l; i < r; ++i) a[i] ^= (1 << k);
  s1 += calc(k, l, r);
  for (int i = l; i < r; ++i) a[i] ^= (1 << k);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", a + i);
  }
  ll ans = 0;
  for (int k = 30; k >= 0; --k) {
    ll s0 = 0, s1 = 0;
    for (int i = 0, j; i < n; i = j) {
      for (j = i; j < n && (a[j] & pm) == (a[i] & pm); ++j)
        ;
      update(k, i, j, s0, s1);
    }
    if (s1 < s0) {
      x |= 1 << k;
      ans += s1;
      for (int i = 0; i < n; ++i) {
        a[i] ^= (1 << k);
      }
    } else {
      ans += s0;
    }
    pm |= 1 << k;
    std::stable_sort(a, a + n,
                     [](int i, int j) { return (i & pm) < (j & pm); });
  }
  printf("%lld %d\n", ans, x);
  return 0;
}

```