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;
template <typename _A, typename _B>
using _p = pair<_A, _B>;
template <typename _A>
using _v = vector<_A>;
template <typename _A>
using _q = queue<_A>;
template <typename _A>
using _s = set<_A>;
template <typename _A>
using _st = stack<_A>;
template <typename _A>
using _d = deque<_A>;
template <typename _A>
using _pq = priority_queue<_A>;
template <typename _A, typename _C>
using _pqc = priority_queue<_A, _v<_A>, _C>;
const int NM = 3e5 + 5;
int N, A[NM];
_v<_v<long long int>> sls(31, _v<long long int>(2, 0));
void parseinds(int bit, _v<int> inds) {
  if (bit < 0 || inds.size() < 2) return;
  _v<long long int> bs(inds.size(), 0);
  bs[0] = !!(A[inds[0]] & 1 << bit);
  for (int i = 1; i < inds.size(); i++)
    bs[i] += bs[i - 1] + !!(A[inds[i]] & 1 << bit);
  for (int i = 1; i < inds.size(); i++) {
    if (A[inds[i]] & 1 << bit)
      sls[bit][1] += i - bs[i - 1];
    else
      sls[bit][0] += bs[i - 1];
  }
  _v<int> zinds, oinds;
  for (int i = 0; i < inds.size(); i++)
    (A[inds[i]] & 1 << bit ? oinds : zinds).push_back(inds[i]);
  parseinds(bit - 1, zinds);
  parseinds(bit - 1, oinds);
}
int tmain(void) {
  cin >> N;
  for (int i = 0; i < N; i++) cin >> A[i];
  _v<int> inds(N);
  iota(inds.begin(), inds.end(), 0);
  parseinds(30, inds);
  long long int ans = 0, x = 0;
  for (int bit = 0; bit < 31; bit++) {
    if (sls[bit][0] > sls[bit][1])
      ans += sls[bit][1], x += 1 << bit;
    else
      ans += sls[bit][0];
  }
  cout << ans << ' ' << x;
  return 0;
}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  tmain();
  return 0;
}
