#include <bits/stdc++.h>
using namespace std;
int i, j, t, n, zero, one, k;
long long d[31][2], res;
struct sb {
  int num;
  bool pre;
} a[300500];
vector<sb> vv;
void dp(int t, vector<sb> v) {
  if (v.size() <= 1 || !t) {
    vector<sb>().swap(v);
    return;
  }
  n = v.size();
  one = zero = 0;
  vector<sb> v0, v1;
  if (v[0].pre) {
    one++;
    v[0].pre = (v[0].num & (1 << (t - 2))) > 0;
    v1.push_back(v[0]);
  } else {
    zero++;
    v[0].pre = (v[0].num & (1 << (t - 2))) > 0;
    v0.push_back(v[0]);
  }
  for (i = 1; i < n; i++) {
    if (v[i].pre) {
      one++;
      d[t][1] += zero;
      v[i].pre = (v[i].num & (1 << (t - 2))) > 0;
      v1.push_back(v[i]);
    } else {
      zero++;
      d[t][0] += one;
      v[i].pre = (v[i].num & (1 << (t - 2))) > 0;
      v0.push_back(v[i]);
    }
  }
  vector<sb>().swap(v);
  dp(t - 1, v0);
  vector<sb>().swap(v0);
  dp(t - 1, v1);
  vector<sb>().swap(v1);
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i].num);
    a[i].pre = (a[i].num & (1 << 29)) > 0;
    vv.push_back(a[i]);
  }
  dp(30, vv);
  for (i = 1; i <= 30; i++) {
    if (d[i][1] < d[i][0]) {
      k += (1 << (i - 1));
      res += d[i][1];
    } else {
      res += d[i][0];
    }
  }
  printf("%lld %d\n", res, k);
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 i, j, t, n, zero, one, k;
long long d[31][2], res;
struct sb {
  int num;
  bool pre;
} a[300500];
vector<sb> vv;
void dp(int t, vector<sb> v) {
  if (v.size() <= 1 || !t) {
    vector<sb>().swap(v);
    return;
  }
  n = v.size();
  one = zero = 0;
  vector<sb> v0, v1;
  if (v[0].pre) {
    one++;
    v[0].pre = (v[0].num & (1 << (t - 2))) > 0;
    v1.push_back(v[0]);
  } else {
    zero++;
    v[0].pre = (v[0].num & (1 << (t - 2))) > 0;
    v0.push_back(v[0]);
  }
  for (i = 1; i < n; i++) {
    if (v[i].pre) {
      one++;
      d[t][1] += zero;
      v[i].pre = (v[i].num & (1 << (t - 2))) > 0;
      v1.push_back(v[i]);
    } else {
      zero++;
      d[t][0] += one;
      v[i].pre = (v[i].num & (1 << (t - 2))) > 0;
      v0.push_back(v[i]);
    }
  }
  vector<sb>().swap(v);
  dp(t - 1, v0);
  vector<sb>().swap(v0);
  dp(t - 1, v1);
  vector<sb>().swap(v1);
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i].num);
    a[i].pre = (a[i].num & (1 << 29)) > 0;
    vv.push_back(a[i]);
  }
  dp(30, vv);
  for (i = 1; i <= 30; i++) {
    if (d[i][1] < d[i][0]) {
      k += (1 << (i - 1));
      res += d[i][1];
    } else {
      res += d[i][0];
    }
  }
  printf("%lld %d\n", res, k);
}

```