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 mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const int maxn = 3e5 + 4;
const int N = 405;
int n, tot = 0;
int trans[maxn * 30][2];
vector<int> ed[maxn * 30];
long long v = 0, x = 0;
long long W[33][2];
void append(int x, int id) {
  int p = 0;
  for (int i = 30; i >= 0; i--) {
    int d = (x >> i) & 1;
    if (!trans[p][d]) trans[p][d] = ++tot;
    p = trans[p][d];
  }
  ed[p].push_back(id);
}
vector<int> dfs(int p, int dep) {
  if (dep == 0) {
    return ed[p];
  }
  vector<int> ls, rs, k;
  if (trans[p][0]) ls = dfs(trans[p][0], dep - 1);
  if (trans[p][1]) rs = dfs(trans[p][1], dep - 1);
  long long c0 = 0;
  for (int i = 0, j = 0; i < ls.size(); i++) {
    while (j < rs.size() && rs[j] < ls[i]) j++;
    c0 += j;
  }
  long long c1 = 0;
  for (int i = 0, j = 0; i < rs.size(); i++) {
    while (j < ls.size() && ls[j] < rs[i]) j++;
    c1 += j;
  }
  W[dep - 1][0] += c0;
  W[dep - 1][1] += c1;
  for (int i = 0; i < ls.size(); i++) k.push_back(ls[i]);
  for (int i = 0; i < rs.size(); i++) k.push_back(rs[i]);
  sort((k).begin(), (k).end());
  return k;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int d;
    scanf("%d", &d);
    append(d, i);
  }
  dfs(0, 31);
  for (int i = 0; i <= 30; i++) {
    if (W[i][0] <= W[i][1]) {
      v = v + W[i][0];
    } else {
      v = v + W[i][1];
      x = x + (1 << i);
    }
  }
  printf("%lld %lld\n", v, x);
  return 0;
}
