#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 3e5 + 5;
int n, ans, a[N], b[N], t[N];
ll Calc(int a[], int l, int r) {
  if (l == r) return 0;
  int mid = (l + r) >> 1, cnt = l, p = l, q = mid + 1;
  ll ret = Calc(a, l, mid) + Calc(a, mid + 1, r);
  while (p <= mid && q <= r)
    if (a[q] < a[p]) {
      t[cnt++] = a[q++];
      ret += mid - p + 1;
    } else
      t[cnt++] = a[p++];
  while (p <= mid) t[cnt++] = a[p++];
  while (q <= r) t[cnt++] = a[q++];
  for (int i = l; i <= r; ++i) a[i] = t[i];
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  for (int j = 29; j >= 0; --j) {
    int t = (1 << 30) - (1 << j);
    for (int i = 1; i <= n; ++i) b[i] = a[i] & t;
    ll x = Calc(b, 1, n);
    for (int i = 1; i <= n; ++i) b[i] = (a[i] & t) ^ (1 << j);
    ll y = Calc(b, 1, n);
    if (y < x) ans += 1 << j;
  }
  for (int i = 1; i <= n; ++i) b[i] = a[i] ^ ans;
  printf("%lld %d\n", Calc(b, 1, n), ans);
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
using ll = long long;
const int N = 3e5 + 5;
int n, ans, a[N], b[N], t[N];
ll Calc(int a[], int l, int r) {
  if (l == r) return 0;
  int mid = (l + r) >> 1, cnt = l, p = l, q = mid + 1;
  ll ret = Calc(a, l, mid) + Calc(a, mid + 1, r);
  while (p <= mid && q <= r)
    if (a[q] < a[p]) {
      t[cnt++] = a[q++];
      ret += mid - p + 1;
    } else
      t[cnt++] = a[p++];
  while (p <= mid) t[cnt++] = a[p++];
  while (q <= r) t[cnt++] = a[q++];
  for (int i = l; i <= r; ++i) a[i] = t[i];
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  for (int j = 29; j >= 0; --j) {
    int t = (1 << 30) - (1 << j);
    for (int i = 1; i <= n; ++i) b[i] = a[i] & t;
    ll x = Calc(b, 1, n);
    for (int i = 1; i <= n; ++i) b[i] = (a[i] & t) ^ (1 << j);
    ll y = Calc(b, 1, n);
    if (y < x) ans += 1 << j;
  }
  for (int i = 1; i <= n; ++i) b[i] = a[i] ^ ans;
  printf("%lld %d\n", Calc(b, 1, n), ans);
  return 0;
}

```