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;
vector<vector<long long>> dp;
void fun(vector<long long> &a, long long bit) {
  if (!a.size() || bit < 0) return;
  long long dp0 = 0, dp1 = 0, z = 0, o = 0;
  vector<long long> l, r;
  for (long long x : a) {
    if (x & (1 << bit)) {
      o++;
      dp1 += z;
      r.push_back(x);
    } else {
      z++;
      dp0 += o;
      l.push_back(x);
    }
  }
  dp[bit][0] += dp0;
  dp[bit][1] += dp1;
  fun(l, bit - 1);
  fun(r, bit - 1);
}
signed main() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; ++i) cin >> a[i];
  dp.resize(32);
  for (long long i = 0; i < 32; ++i) {
    dp[i].push_back(0);
    dp[i].push_back(0);
  }
  fun(a, 30);
  for (long long i = 0; i < 32; ++i) {
    cerr << dp[i][0] << " " << dp[i][1] << endl;
  }
  long long ans = 0, val = 0;
  for (long long i = 0; i < 32; ++i) {
    if (dp[i][0] <= dp[i][1])
      ans += dp[i][0];
    else {
      ans += dp[i][1];
      val |= (1 << i);
    }
  }
  cout << ans << " " << val;
  return 0;
}
