#include <bits/stdc++.h>
using namespace std;
int n;
long long *a;
long long ans, tz;
long long su[2][64];
void calc(long long *a, int len, int now) {
  if (!len) {
    free(a);
    return;
  }
  if (now < 0) {
    free(a);
    return;
  }
  long long temp[2][2];
  for (int i = 0; i < 2; i++)
    for (int j = 0; j < 2; j++) temp[i][j] = 0;
  for (int i = 0; i < len; i++) {
    long long tz = a[i] >> now;
    temp[0][tz % 2] += temp[1][(tz + 1) % 2];
    temp[1][tz % 2]++;
  }
  su[1][now] += temp[0][0];
  su[0][now] += temp[0][1];
  long long *b, *c;
  b = (long long *)malloc(temp[1][0] * sizeof(long long));
  c = (long long *)malloc(temp[1][1] * sizeof(long long));
  int cot[2];
  cot[0] = cot[1] = 0;
  for (int i = 0; i < len; i++) {
    long long tz = a[i] >> now;
    if (tz % 2) {
      c[cot[1]++] = a[i];
    } else {
      b[cot[0]++] = a[i];
    }
  }
  free(a);
  calc(b, cot[0], now - 1);
  calc(c, cot[1], now - 1);
}
int main() {
  scanf("%d", &n);
  a = (long long *)malloc(n * sizeof(long long));
  for (int i = 0; i < n; i++) scanf("%lld", &a[i]);
  calc(a, n, 35);
  ans = 0;
  for (int i = 0; i < 35; i++) {
    tz += min(su[1][i], su[0][i]);
    if (su[1][i] > su[0][i]) ans += ((long long)1 << i);
  }
  printf("%lld %lld", tz, ans);
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int n;
long long *a;
long long ans, tz;
long long su[2][64];
void calc(long long *a, int len, int now) {
  if (!len) {
    free(a);
    return;
  }
  if (now < 0) {
    free(a);
    return;
  }
  long long temp[2][2];
  for (int i = 0; i < 2; i++)
    for (int j = 0; j < 2; j++) temp[i][j] = 0;
  for (int i = 0; i < len; i++) {
    long long tz = a[i] >> now;
    temp[0][tz % 2] += temp[1][(tz + 1) % 2];
    temp[1][tz % 2]++;
  }
  su[1][now] += temp[0][0];
  su[0][now] += temp[0][1];
  long long *b, *c;
  b = (long long *)malloc(temp[1][0] * sizeof(long long));
  c = (long long *)malloc(temp[1][1] * sizeof(long long));
  int cot[2];
  cot[0] = cot[1] = 0;
  for (int i = 0; i < len; i++) {
    long long tz = a[i] >> now;
    if (tz % 2) {
      c[cot[1]++] = a[i];
    } else {
      b[cot[0]++] = a[i];
    }
  }
  free(a);
  calc(b, cot[0], now - 1);
  calc(c, cot[1], now - 1);
}
int main() {
  scanf("%d", &n);
  a = (long long *)malloc(n * sizeof(long long));
  for (int i = 0; i < n; i++) scanf("%lld", &a[i]);
  calc(a, n, 35);
  ans = 0;
  for (int i = 0; i < 35; i++) {
    tz += min(su[1][i], su[0][i]);
    if (su[1][i] > su[0][i]) ans += ((long long)1 << i);
  }
  printf("%lld %lld", tz, ans);
}

```