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 int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const long long INF = 1e18;
const int M = 3e5 + 5;
int tr[M * 29][2];
long long dp[100][2];
int cnt;
vector<int> g[M * 29];
void pn() { cout << "NO" << '\n'; }
void py() { cout << "YES" << '\n'; }
void add(int x, int pos) {
  int rt = 0;
  for (int i = 29; i >= 0; i--) {
    int v = ((x >> i) & 1);
    if (!tr[rt][v]) tr[rt][v] = ++cnt;
    rt = tr[rt][v];
    g[rt].push_back(pos);
  }
}
void dfs(int rt, int now) {
  int l = tr[rt][0], r = tr[rt][1];
  if (l) dfs(l, now - 1);
  if (r) dfs(r, now - 1);
  if (!l || !r) return;
  int i = 0;
  long long res = 0;
  for (int j = 0; j < (int)g[l].size(); j++) {
    while (i < (int)g[r].size() && g[l][j] > g[r][i]) i++;
    res += i;
  }
  dp[now][0] += res;
  dp[now][1] += (long long)g[l].size() * (long long)g[r].size() - res;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int x, i = 1; i <= n; i++) {
    scanf("%d", &x);
    add(x, i);
  }
  dfs(0, 29);
  long long sum = 0, x = 0;
  for (int i = 29; i >= 0; i--) {
    if (dp[i][0] > dp[i][1]) x += (1 << i);
    sum += min(dp[i][0], dp[i][1]);
  }
  printf("%lld %lld\n", sum, x);
  return 0;
}
