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 N = 3e5 + 100;
int trie[N * 31][2], tot;
vector<int> g[N * 31];
long long dp[31][2];
void insert(int x, int id) {
  int now = 0;
  for (int i = 30; i >= 0; i--) {
    int c = (x >> i & 1);
    if (!trie[now][c]) trie[now][c] = ++tot;
    now = trie[now][c];
    g[now].push_back(id);
  }
}
void dfs(int dep, int u) {
  int ls = trie[u][0], rs = trie[u][1];
  if (ls) dfs(dep - 1, ls);
  if (rs) dfs(dep - 1, rs);
  if (!ls || !rs) return;
  int j = 0;
  long long sum = 0;
  for (int i = 0; i < (int)g[ls].size(); i++) {
    while (j < (int)g[rs].size() && g[rs][j] < g[ls][i]) j++;
    sum += j;
  }
  dp[dep][0] += sum;
  dp[dep][1] += 1ll * g[rs].size() * g[ls].size() - sum;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    insert(x, i);
  }
  dfs(30, 0);
  long long ans = 0, ret = 0;
  for (int i = 30; i >= 0; i--) {
    ans += min(dp[i][0], dp[i][1]);
    if (dp[i][1] < dp[i][0]) {
      ret += (1ll << i);
    }
  }
  printf("%lld %lld\n", ans, ret);
  return 0;
}
