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;
#pragma GCC optimize("-O3")
vector<long long> v[8000000];
long long isithBitSet(long long n, long long k) {
  if (n & (1LL << (k)))
    return 1;
  else
    return 0;
}
vector<long long> tempOne;
vector<long long> tempZero;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  long long x;
  long long index = -1;
  vector<long long> finalAns;
  for (long long i = 0; i < n; i++) {
    cin >> x;
    v[index + 1].push_back(x);
  }
  long long ans = 0;
  long long count = 0;
  for (long long i = 31; i >= 0; i--) {
    long long temp = index + 1;
    index = count;
    long long ansOne = 0;
    long long ansZero = 0;
    for (long long j = temp; j <= index; j++) {
      long long len = v[j].size();
      long long one = 0;
      long long zero = 0;
      tempOne.clear();
      tempZero.clear();
      for (long long k = (len - 1); k >= 0; k--) {
        if (isithBitSet(v[j][k], i)) {
          ansZero += zero;
          one++;
          tempZero.push_back(v[j][k]);
        } else {
          ansOne += one;
          zero++;
          tempOne.push_back(v[j][k]);
        }
      }
      if (tempOne.size() > 1) {
        v[count + 1] = tempOne;
        reverse((v[count + 1]).begin(), (v[count + 1]).end());
        count++;
      }
      if (tempZero.size() > 1) {
        v[count + 1] = tempZero;
        reverse((v[count + 1]).begin(), (v[count + 1]).end());
        count++;
      }
    }
    ans += min(ansOne, ansZero);
    if (ansZero <= ansOne) {
      finalAns.push_back(0);
    } else {
      finalAns.push_back(1);
    }
  }
  cout << ans << " ";
  ans = 0;
  long long power2 = 1;
  reverse((finalAns).begin(), (finalAns).end());
  for (long long i = 0; i < (long long)finalAns.size(); i++) {
    ans += finalAns[i] * power2;
    power2 *= 2;
  }
  cout << ans;
  return 0;
}
