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;
int n, tri[8000010][2], tot;
long long f[31][2];
vector<int> son[8000010];
inline void insert(int x, int id) {
  int rt = 0;
  for (int i = 29; i >= 0; --i) {
    int v = (x >> i) & 1;
    if (!tri[rt][v]) tri[rt][v] = ++tot;
    rt = tri[rt][v];
    son[rt].push_back(id);
  }
}
void dp(int rt, int bit) {
  int l = tri[rt][0], r = tri[rt][1];
  if (l) dp(l, bit - 1);
  if (r) dp(r, bit - 1);
  if (!l || !r) return;
  long long cnt = 0, ptr = 0;
  for (int i = 0; i < son[l].size(); ++i) {
    while (ptr < son[r].size() && son[r][ptr] < son[l][i]) ++ptr;
    cnt += ptr;
  }
  f[bit][0] += cnt, f[bit][1] += (long long)son[l].size() * son[r].size() - cnt;
}
signed main() {
  cin >> n;
  for (int i = 1, a; i <= n; ++i) {
    cin >> a;
    insert(a, i);
  }
  dp(0, 29);
  long long cnt = 0, ans = 0;
  for (int i = 0; i <= 29; ++i) {
    cnt += min(f[i][1], f[i][0]);
    if (f[i][1] < f[i][0]) ans |= (1 << i);
  }
  return cout << cnt << ' ' << ans, 0;
}
