#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e6;
const int L = 30;
int N;
int a[MAXN];
long long cnt[L + 1][2];
void f(int st, int en, int d) {
  if (st == en || d < 0) {
    return;
  }
  vector<int> v[2];
  long long inv = 0;
  for (int i = st; i < en; i++) {
    int x = (a[i] >> d) & 1;
    v[x].push_back(a[i]);
    if (x == 0) {
      inv += v[1].size();
    }
  }
  int n0 = v[0].size(), n1 = v[1].size();
  cnt[d][0] += inv;
  cnt[d][1] += 1ll * n0 * n1 - inv;
  for (int i = 0; i < n0; i++) {
    a[st + i] = v[0][i];
  }
  for (int i = 0; i < n1; i++) {
    a[st + n0 + i] = v[1][i];
  }
  f(st, st + n0, d - 1);
  f(st + n0, en, d - 1);
}
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; i++) {
    scanf("%d", &a[i]);
  }
  f(0, N, L);
  long long ret = 0;
  int x = 0;
  for (int i = 0; i <= L; i++) {
    if (cnt[i][1] < cnt[i][0]) {
      x |= 1 << i;
      ret += cnt[i][1];
    } else {
      ret += cnt[i][0];
    }
  }
  printf("%lld %d\n", ret, x);
}

### 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>
using namespace std;
const int MAXN = 1e6;
const int L = 30;
int N;
int a[MAXN];
long long cnt[L + 1][2];
void f(int st, int en, int d) {
  if (st == en || d < 0) {
    return;
  }
  vector<int> v[2];
  long long inv = 0;
  for (int i = st; i < en; i++) {
    int x = (a[i] >> d) & 1;
    v[x].push_back(a[i]);
    if (x == 0) {
      inv += v[1].size();
    }
  }
  int n0 = v[0].size(), n1 = v[1].size();
  cnt[d][0] += inv;
  cnt[d][1] += 1ll * n0 * n1 - inv;
  for (int i = 0; i < n0; i++) {
    a[st + i] = v[0][i];
  }
  for (int i = 0; i < n1; i++) {
    a[st + n0 + i] = v[1][i];
  }
  f(st, st + n0, d - 1);
  f(st + n0, en, d - 1);
}
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; i++) {
    scanf("%d", &a[i]);
  }
  f(0, N, L);
  long long ret = 0;
  int x = 0;
  for (int i = 0; i <= L; i++) {
    if (cnt[i][1] < cnt[i][0]) {
      x |= 1 << i;
      ret += cnt[i][1];
    } else {
      ret += cnt[i][0];
    }
  }
  printf("%lld %d\n", ret, x);
}

```