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 mod = 1e9 + 7;
const long long INF = 2e18 + 1;
const int inf = 1e9 + 5;
const double eps = 1e-8;
const int $ = 4e6 + 5;
vector<int> v[$];
long long dp[33][2];
int rt[$][2];
int tot = 0;
void add(long long x, int y) {
  int now = 0;
  for (int i = 30; i >= 0; i--) {
    int p = (x >> i) & 1;
    if (!rt[now][p]) rt[now][p] = ++tot;
    now = rt[now][p];
    v[now].push_back(y);
  }
}
void dfs(int num, int i) {
  if (i == -1) return;
  int l = rt[num][0], r = rt[num][1];
  long long sl = v[l].size(), sr = v[r].size();
  long long res = 0, now = 0;
  for (auto it : v[l]) {
    if (!sr) break;
    while (now < sr && v[r][now] < it) now++;
    res += now;
  }
  dp[i][0] += res;
  dp[i][1] += sl * sr - res;
  if (rt[num][0]) dfs(rt[num][0], i - 1);
  if (rt[num][1]) dfs(rt[num][1], i - 1);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  long long x, ans = 0, num = 0;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    add(x, i);
  }
  dfs(0, 30);
  for (int i = 0; i <= 30; i++) {
    num += min(dp[i][0], dp[i][1]);
    if (dp[i][0] > dp[i][1]) ans += (1 << i);
  }
  cout << num << ' ' << ans << endl;
  return 0;
}
