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;
const long long base = 36;
const long long base2 = 316;
const long long mod = 1e9 + 7;
const long long mod2 = 1e9 + 9;
const long long INF = 1e18;
long long n;
long long a[300005];
long long sum[31][2];
struct trieNode {
  trieNode *child[2];
  vector<long long> hold;
  trieNode() {
    hold.clear();
    child[0] = NULL;
    child[1] = NULL;
  }
};
trieNode *root = new trieNode;
void Insert(long long x) {
  trieNode *p = root;
  for (long long i = 30; i >= 0; i--) {
    long long bit = ((a[x] >> i) & 1);
    if (p->child[bit] == NULL) p->child[bit] = new trieNode;
    p = p->child[bit];
    p->hold.push_back(x);
  }
}
void calc(trieNode *p, long long k) {
  if (p->child[0] != NULL) calc(p->child[0], k - 1);
  if (p->child[1] != NULL) calc(p->child[1], k - 1);
  if (p->child[0] == NULL || p->child[1] == NULL) return;
  vector<long long> x = p->child[0]->hold, y = p->child[1]->hold;
  long long cnt = 0, tmp = 0;
  for (auto v : x) {
    while (cnt < y.size() && (v > y[cnt])) cnt++;
    tmp += cnt;
  }
  sum[k][0] += tmp;
  sum[k][1] += (long long)x.size() * (long long)y.size() - tmp;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    Insert(i);
  }
  calc(root, 30);
  long long cnt = 0, ans = 0;
  for (long long i = 30; i >= 0; i--) {
    if (sum[i][1] < sum[i][0]) {
      cnt += sum[i][1];
      ans |= (1 << i);
    } else
      cnt += sum[i][0];
  }
  cout << cnt << " " << ans;
}
