#include <bits/stdc++.h>
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
inline int read() {
  register int res = 0;
  register char c = getchar(), f = 1;
  while (c < 48 || c > 57) {
    if (c == '-') f = 0;
    c = getchar();
  }
  while (c >= 48 && c <= 57)
    res = (res << 3) + (res << 1) + (c & 15), c = getchar();
  return f ? res : -res;
}
inline void write(long long x) {
  register char c[21], len = 0;
  if (!x) return putchar('0'), void();
  if (x < 0) x = -x, putchar('-');
  while (x) c[++len] = x % 10, x /= 10;
  while (len) putchar(c[len--] + 48);
}
const int N = 3e5 + 10;
int n, x, a[N];
long long ans;
struct node {
  int c0, c1;
  long long s0, s1;
};
unordered_map<int, node> mp;
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  for (int i = 29; ~i; --i) {
    mp.clear();
    for (int j = 1, o, t; j <= n; ++j) {
      o = a[j] >> i & 1, t = a[j] >> (i + 1);
      o ? (mp[t].s1 += mp[t].c0, ++mp[t].c1)
        : (mp[t].s0 += mp[t].c1, ++mp[t].c0);
    }
    long long s0 = 0, s1 = 0;
    for (auto p : mp) s0 += p.second.s0, s1 += p.second.s1;
    if (s0 > s1)
      x |= 1 << i, ans += s1;
    else
      ans += s0;
  }
  write(ans), putchar(' '), write(x), putchar('\n');
}

### 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>
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
inline int read() {
  register int res = 0;
  register char c = getchar(), f = 1;
  while (c < 48 || c > 57) {
    if (c == '-') f = 0;
    c = getchar();
  }
  while (c >= 48 && c <= 57)
    res = (res << 3) + (res << 1) + (c & 15), c = getchar();
  return f ? res : -res;
}
inline void write(long long x) {
  register char c[21], len = 0;
  if (!x) return putchar('0'), void();
  if (x < 0) x = -x, putchar('-');
  while (x) c[++len] = x % 10, x /= 10;
  while (len) putchar(c[len--] + 48);
}
const int N = 3e5 + 10;
int n, x, a[N];
long long ans;
struct node {
  int c0, c1;
  long long s0, s1;
};
unordered_map<int, node> mp;
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  for (int i = 29; ~i; --i) {
    mp.clear();
    for (int j = 1, o, t; j <= n; ++j) {
      o = a[j] >> i & 1, t = a[j] >> (i + 1);
      o ? (mp[t].s1 += mp[t].c0, ++mp[t].c1)
        : (mp[t].s0 += mp[t].c1, ++mp[t].c0);
    }
    long long s0 = 0, s1 = 0;
    for (auto p : mp) s0 += p.second.s0, s1 += p.second.s1;
    if (s0 > s1)
      x |= 1 << i, ans += s1;
    else
      ans += s0;
  }
  write(ans), putchar(' '), write(x), putchar('\n');
}

```