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;
long long n, a[300015], b[300015], res[32][2];
void solve(long long l, long long r, long long dep) {
  if (l > r || dep < 0) return;
  long long v = (1 << dep), one, zero, ao, az;
  one = zero = ao = az = 0;
  for (long long i = l; i <= r; i++) {
    if (a[i] & v)
      one++, ao += zero;
    else
      zero++, az += one;
  }
  res[dep][0] += az;
  res[dep][1] += ao;
  long long ll = l, rr = l + one;
  for (long long i = l; i <= r; i++) {
    if (a[i] & v)
      b[ll++] = a[i];
    else
      b[rr++] = a[i];
  }
  for (long long i = l; i <= r; i++) a[i] = b[i];
  solve(l, ll - 1, dep - 1);
  solve(ll, r, dep - 1);
}
signed main() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
  solve(1, n, 30);
  long long x, ans;
  x = ans = 0;
  for (long long i = 0; i <= 30; i++) {
    if (res[i][0] > res[i][1]) x += (1 << i);
    ans += min(res[i][0], res[i][1]);
  }
  printf("%lld %lld\n", ans, x);
  return 0;
}
