#include <bits/stdc++.h>
#pragma GCC optimize(2)
using namespace std;
inline int read() {
  int x = 0;
  int f = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    f |= (ch == '-') ? 1 : 0;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return f ? -x : x;
}
const int maxn = 3e5 + 10;
long long ans[maxn][2];
int aa[maxn];
int tmp1[maxn];
int tmp2[maxn];
int n;
inline void divi(int l, int r, int k) {
  if (l >= r || k < 0) return;
  int num1 = 0, num2 = 0;
  for (register int i = l; i <= r; ++i) {
    if ((aa[i] >> k) & 1) {
      tmp1[++num1] = aa[i];
      ans[k][1] += num2;
    } else {
      tmp2[++num2] = aa[i];
      ans[k][0] += num1;
    }
  }
  for (register int i = l; i <= l + num1 - 1; ++i) {
    aa[i] = tmp1[i - l + 1];
  }
  for (register int i = l + num1; i <= r; ++i) aa[i] = tmp2[i - l - num1 + 1];
  divi(l, l + num1 - 1, k - 1);
  divi(l + num1, r, k - 1);
}
int main() {
  n = read();
  for (register int i = 1; i <= n; ++i) aa[i] = read();
  divi(1, n, 31);
  long long res1 = 0, res2 = 0;
  for (register int i = 0; i <= 31; ++i) {
    if (ans[i][0] > ans[i][1]) res2 += (1 << i);
    res1 += min(ans[i][0], ans[i][1]);
  }
  printf("%lld %lld", res1, res2);
}

### Prompt

In CPP, your task is to solve 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>
#pragma GCC optimize(2)
using namespace std;
inline int read() {
  int x = 0;
  int f = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    f |= (ch == '-') ? 1 : 0;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return f ? -x : x;
}
const int maxn = 3e5 + 10;
long long ans[maxn][2];
int aa[maxn];
int tmp1[maxn];
int tmp2[maxn];
int n;
inline void divi(int l, int r, int k) {
  if (l >= r || k < 0) return;
  int num1 = 0, num2 = 0;
  for (register int i = l; i <= r; ++i) {
    if ((aa[i] >> k) & 1) {
      tmp1[++num1] = aa[i];
      ans[k][1] += num2;
    } else {
      tmp2[++num2] = aa[i];
      ans[k][0] += num1;
    }
  }
  for (register int i = l; i <= l + num1 - 1; ++i) {
    aa[i] = tmp1[i - l + 1];
  }
  for (register int i = l + num1; i <= r; ++i) aa[i] = tmp2[i - l - num1 + 1];
  divi(l, l + num1 - 1, k - 1);
  divi(l + num1, r, k - 1);
}
int main() {
  n = read();
  for (register int i = 1; i <= n; ++i) aa[i] = read();
  divi(1, n, 31);
  long long res1 = 0, res2 = 0;
  for (register int i = 0; i <= 31; ++i) {
    if (ans[i][0] > ans[i][1]) res2 += (1 << i);
    res1 += min(ans[i][0], ans[i][1]);
  }
  printf("%lld %lld", res1, res2);
}

```