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.
#include <bits/stdc++.h>
using namespace std;
int m[31], mask = 0;
struct num {
  int v, idx;
  bool operator<(const num &o) const {
    if ((v & mask) != (o.v & mask)) return (v & mask) < (o.v & mask);
    return idx < o.idx;
  }
};
num a[300001];
int main() {
  m[1] = 1;
  for (int i = 2; i <= 30; ++i) m[i] = m[i - 1] * 2;
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i].v), a[i].idx = i;
  long long rescnt = 0;
  int resnum = 0;
  mask = 0;
  for (int i = 30; i >= 1; --i) {
    sort(a, a + n);
    long long zero = 0, one = 0;
    long long cnt_zero = 0, cnt_one = 0;
    for (int j = 0; j < n; ++j) {
      if (j != 0 && (a[j].v & mask) != (a[j - 1].v & mask)) {
        cnt_zero = 0;
        cnt_one = 0;
      }
      if ((a[j].v & m[i]) == 0) {
        ++cnt_one;
        zero += cnt_zero;
      } else {
        ++cnt_zero;
        one += cnt_one;
      }
    }
    if (one < zero) {
      rescnt += one;
      resnum += m[i];
    } else {
      rescnt += zero;
    }
    mask += m[i];
  }
  printf("%lld %d\n", rescnt, resnum);
}
