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;
const int N = 5e6 + 5, M = 2e4 + 5, inf = 0x3f3f3f3f, mod = 1e9 + 7;
int ch[N][2], sz[N], id, n;
long long dp[30][2];
void add(int x) {
  int p = 0;
  for (int i = 29; ~i; i--) {
    int v = (x >> i) & 1;
    if (!ch[p][v]) ch[p][v] = ++id;
    dp[i][v ^ 1] += sz[ch[p][v ^ 1]];
    p = ch[p][v], sz[p]++;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    add(x);
  }
  long long cnt = 0, ans = 0;
  for (int i = 29; ~i; i--) {
    if (dp[i][0] < dp[i][1])
      ans += (1 << i), cnt += dp[i][0];
    else
      cnt += dp[i][1];
  }
  printf("%lld %lld\n", cnt, ans);
  return 0;
}
