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 long long mod = 2147493647;
int t, n;
int a[1000010];
int bit[1000010][34];
long long c[34][2];
long long ans, x;
bool flag[1000010];
void solve(int pos, vector<int> vec) {
  if (pos == 0 || vec.empty()) return;
  int cnt0 = 0, cnt1 = 0;
  long long tot0 = 0, tot1 = 0;
  vector<int> vecl, vecr;
  for (int j = 0; j < vec.size(); ++j) {
    if (bit[vec[j]][pos] == 0) {
      vecl.push_back(vec[j]);
      cnt0++;
      tot0 += cnt1;
    } else {
      vecr.push_back(vec[j]);
      cnt1++;
      tot1 += cnt0;
    }
  }
  c[pos][0] += tot0;
  c[pos][1] += tot1;
  solve(pos - 1, vecl);
  solve(pos - 1, vecr);
}
int main() {
  t = 1;
  while (t--) {
    vector<int> vv;
    x = 0;
    ans = 0;
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
      scanf("%d", &a[i]);
      int tmp = a[i], tot = 0;
      while (tmp) {
        bit[i][++tot] = tmp % 2;
        tmp /= 2;
      }
      vv.push_back(i);
    }
    solve(32, vv);
    for (int i = 1; i <= 32; ++i) {
      if (c[i][0] > c[i][1]) {
        x += (1 << (i - 1));
        ans += c[i][1];
      } else
        ans += c[i][0];
    }
    printf("%lld %lld\n", ans, x);
  }
  return 0;
}
