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>
#pragma GCC target("avx")
#pragma GCC optimize(2)
#pragma GCC optimize(3, "Ofast", "inline")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
const int N = 1e6 + 10, M = 5e5 + 10;
const int mod = 1e9 + 7;
int head[M], num_edge;
struct Edge {
  int to;
  int Next;
  int dis;
} edge[M];
void add_edge(int from, int to, int dis) {
  edge[++num_edge].to = to;
  edge[num_edge].dis = dis;
  edge[num_edge].Next = head[from];
  head[from] = num_edge;
}
int prime[N + 10], vis[N + 10], cnt;
void init() {
  for (int i = 2; i <= N; i++) {
    if (!vis[i]) prime[++cnt] = i;
    for (int j = 1; i * prime[j] <= N && j <= cnt; j++) {
      vis[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}
long long quick_pow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
vector<long long> G[300010], dp[5];
long long a[300010];
int main() {
  long long n, ans = 0, cnt = 0, tot = 1;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    G[1].push_back(a[i]);
  }
  for (int i = 30; i >= 0; i--) {
    long long fir = 0;
    long long sec = 0;
    long long now = tot;
    for (int j = 1; j <= now; j++) {
      for (int i = 0; i < 2; i++) dp[i].clear();
      int sz = G[j].size();
      for (int k = 0; k < sz; k++)
        dp[(G[j][k] >> i + 1) & 1].push_back(G[j][k]);
      if (dp[0].size() != 0 && dp[1].size() != 0) {
        G[j] = dp[1];
        G[++tot] = dp[0];
      } else {
        if (dp[1].size() != 0)
          G[j] = dp[1];
        else
          G[j] = dp[0];
      }
    }
    int X, Y;
    for (int j = 1; j <= tot; j++) {
      X = 0;
      Y = 0;
      int sz = G[j].size();
      for (int k = 0; k < sz; k++) {
        if (((G[j][k] >> i) & 1)) {
          Y++;
          sec = sec + X;
        } else {
          X++;
          fir = fir + Y;
        }
      }
    }
    if (fir > sec) {
      ans = ans + (1 << i);
      cnt = cnt + sec;
    } else
      cnt = cnt + fir;
  }
  printf("%lld %lld\n", cnt, ans);
  return 0;
}
