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;
int aa[2][610000], n, shit, mid;
long long to1, to2;
inline void chuli(int *a, int top, int *b) {
  int zt = 0, ot = 0;
  for (int i = top; i >= 1; i--) {
    if (a[i] & shit)
      ot++;
    else
      zt++;
  }
  int zc = 0, oc = 0;
  for (int i = top; i >= 1; i--) {
    if (a[i] & shit) {
      oc++;
      b[top - oc + 1] = a[i];
      to2 += zc;
    } else {
      zc++;
      b[zt - zc + 1] = a[i];
      to1 += oc;
    }
  }
  mid = zt;
}
int ll[2][610000], rr[2][610000], top[2];
int main() {
  long long ans2 = 0, ans1 = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &aa[0][i]);
  top[0] = 1;
  ll[0][1] = 1;
  rr[0][1] = n;
  int pre = 1, now = 0;
  for (int i = 30; i >= 1; i--) {
    now ^= 1;
    pre ^= 1;
    shit = (1 << (i - 1));
    top[now] = 0;
    to1 = to2 = 0;
    for (int i = 1; i <= top[pre]; i++) {
      chuli(aa[pre] + ll[pre][i] - 1, rr[pre][i] - ll[pre][i] + 1,
            aa[now] + ll[pre][i] - 1);
      if (mid > 0) {
        int x = ++top[now];
        ll[now][x] = ll[pre][i];
        rr[now][x] = ll[pre][i] + mid - 1;
      }
      if (mid != rr[pre][i] - ll[pre][i] + 1) {
        int x = ++top[now];
        ll[now][x] = ll[pre][i] + mid;
        rr[now][x] = rr[pre][i];
      }
    }
    if (to1 >= to2)
      ans2 += to2;
    else
      ans2 += to1, ans1 ^= shit;
  }
  printf("%lld %lld\n", ans2, ans1);
  return 0;
}
