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;
void cetak(const vector<int>& a) {
  for (int i = 0; i < a.size(); i++) {
    printf("%d ", a[i]);
  }
  printf("\n");
}
const int maxn = 5000001;
int t[maxn][2];
int a[300001];
vector<int> indices[maxn];
long long numInv[maxn][2];
int idTrie;
void addNum(int num, int idx) {
  int height = 0;
  for (int i = 29; i >= 0; i--) {
    int digit = (num >> i) & 1;
    if (t[height][digit] == -1) {
      t[height][digit] = ++idTrie;
    }
    height = t[height][digit];
    indices[height].push_back(idx);
  }
}
void go(int height, int b = 29) {
  int l = t[height][0], r = t[height][1];
  if (l != -1) {
    go(l, b - 1);
  }
  if (r != -1) {
    go(r, b - 1);
  }
  if (l == -1 || r == -1) {
    return;
  }
  long long ans = 0;
  int index = 0;
  for (int idxKecil : indices[l]) {
    while (index < indices[r].size() && indices[r][index] < idxKecil) {
      index++;
    }
    ans += index;
  }
  numInv[b][0] += ans;
  numInv[b][1] += ((long long)indices[r].size() * indices[l].size()) - ans;
}
int main() {
  int n, i, j;
  while (scanf("%d", &n) != EOF) {
    idTrie = 0;
    for (i = 0; i < maxn; i++) {
      indices[i].clear();
      for (j = 0; j < 2; j++) {
        t[i][j] = -1;
        numInv[i][j] = 0;
      }
    }
    for (i = 0; i < n; i++) {
      scanf("%d", &a[i]);
      addNum(a[i], i);
    }
    go(0);
    long long jawabInv = 0;
    int sor = 0;
    for (i = 0; i <= 29; i++) {
      long long minim = min(numInv[i][0], numInv[i][1]);
      if (numInv[i][1] < numInv[i][0]) {
        sor += (1 << i);
      }
      jawabInv += minim;
    }
    printf("%lld %d\n", jawabInv, sor);
  }
  return 0;
};
