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>
const int N = 300001;
using namespace std;
string a[N];
int xr = 0;
long long int inv = 0;
void f(vector<vector<int>> &b, int bit) {
  if (bit == -1) return;
  long long int fir = 0;
  vector<vector<int>> nex;
  for (auto &bb : b) {
    int one = 0;
    vector<int> zv, ov;
    for (int ind : bb) {
      if (a[ind][bit] == '1') {
        one++;
        ov.push_back(ind);
      } else {
        fir += one;
        zv.push_back(ind);
      }
    }
    if (!zv.empty()) nex.push_back(zv);
    if (!ov.empty()) nex.push_back(ov);
  }
  long long int sec = 0;
  for (auto &bb : b) {
    int one = 0;
    for (int ind : bb) {
      if (a[ind][bit] == '0') {
        one++;
      } else {
        sec += one;
      }
    }
  }
  if (fir <= sec) {
    inv += fir;
  } else {
    inv += sec;
    xr += 1 << bit;
  }
  f(nex, bit - 1);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  int mb = 30;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    a[i] = string(mb, '0');
    for (int j = 0; j < mb; j++) {
      a[i][j] = (x & (1 << j) ? '1' : '0');
    }
  }
  vector<int> v(n);
  for (int i = 0; i < n; i++) v[i] = i;
  vector<vector<int>> j = {v};
  f(j, mb - 1);
  cout << inv << ' ' << xr;
}
