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 optimize("Ofast", "unroll-loops")
using namespace std;
const long long inf = 1ll << 60;
const long long iinf = 2147483647;
const long long mod = 1e9 + 7;
const long long maxn = 1e6 + 5;
const double PI = acos(-1);
long long pw(long long x, long long p) {
  long long ret = 1;
  while (p > 0) {
    if (p & 1) {
      ret *= x;
      ret %= mod;
    }
    x *= x;
    x %= mod;
    p >>= 1;
  }
  return ret;
}
long long inv(long long a) { return pw(a, mod - 2); }
vector<string> vc;
vector<long long> ans0(maxn), ans1(maxn);
void run(long long bpos, vector<long long> &id) {
  if (bpos == -1 || (long long)(id.size()) == 0) return;
  long long res = 0, cnt = 0, rres = 0, rcnt = 0;
  vector<long long> a, b;
  for (long long i = ((long long)(id.size())) - 1; i >= 0; i--) {
    if (vc[id[i]][bpos] == '0') {
      a.push_back(id[i]);
    } else {
      b.push_back(id[i]);
    }
    if (i == (long long)(id.size()) - 1) {
      if (vc[id[i]][bpos] - '0')
        rcnt++;
      else
        cnt++;
      continue;
    }
    if (vc[id[i]][bpos] == '0') {
      rres += rcnt;
      cnt++;
    } else {
      res += cnt;
      rcnt++;
    }
  }
  ans0[bpos] += res;
  ans1[bpos] += rres;
  reverse(a.begin(), a.end());
  reverse(b.begin(), b.end());
  run(bpos - 1, a);
  run(bpos - 1, b);
}
signed main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long n;
  cin >> n;
  vc = vector<string>(n);
  for (long long i = (0); i < (n); i++) {
    long long x;
    cin >> x;
    for (long long j = (0); j < (31); j++) {
      vc[i] += (((1 << j) & x) > 0) + '0';
    }
  }
  vector<long long> tmp(n);
  for (long long i = (0); i < (n); i++) tmp[i] = i;
  run(30, tmp);
  long long btres = 0, x = 0;
  for (long long i = (0); i < (31); i++) {
    if (ans1[i] < ans0[i]) {
      x += (1 << i);
      btres += ans1[i];
    } else {
      btres += ans0[i];
    }
  }
  cout << btres << ' ' << x << '\n';
}
