#include <bits/stdc++.h>
using namespace std;
struct P {
  int x, y;
  bool operator<(const P &a) const {
    if (y != a.y) return y > a.y;
    return x < a.x;
  };
};
int a, b, c, d, i, n, m, e, k, dx[10] = {1, 0, -1, 0, 1, 1, -1, -1},
                               dy[10] = {0, 1, 0, -1, 1, -1, 1, -1};
int o[333331];
int l[333331][3];
int ss[1];
long long x, y, z, mod = 1000000007, hf = (mod + 1) / 2;
char r[1];
vector<int> v[3];
P u[333331];
set<int> s;
map<int, int> p;
bool as(P a, P b) {
  if (a.x / e != b.x / e) return a.x / e < b.x / e;
  return a.y < b.y;
}
int main() {
  scanf("%d", &a);
  for (int t = 1; t <= a; u[t] = {n, t}, t++) scanf("%d", &n);
  for (int t = 1; t <= 30; t++) {
    e = (1 << (t));
    sort(u + 1, u + a + 1, as);
    n = m = 0;
    x = y = 0;
    for (int i = 1; i <= a; i++) {
      if (u[i - 1].x / e != u[i].x / e) {
        n = m = 0;
      }
      if (u[i].x % e >= e / 2)
        y += n, m++;
      else
        x += m, n++;
    }
    if (x > y) {
      for (int t = 1; t <= a; t++) u[t].x ^= (e / 2);
      z += y, k += e / 2;
    } else
      z += x;
  }
  printf("%lld %d", z, 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;
struct P {
  int x, y;
  bool operator<(const P &a) const {
    if (y != a.y) return y > a.y;
    return x < a.x;
  };
};
int a, b, c, d, i, n, m, e, k, dx[10] = {1, 0, -1, 0, 1, 1, -1, -1},
                               dy[10] = {0, 1, 0, -1, 1, -1, 1, -1};
int o[333331];
int l[333331][3];
int ss[1];
long long x, y, z, mod = 1000000007, hf = (mod + 1) / 2;
char r[1];
vector<int> v[3];
P u[333331];
set<int> s;
map<int, int> p;
bool as(P a, P b) {
  if (a.x / e != b.x / e) return a.x / e < b.x / e;
  return a.y < b.y;
}
int main() {
  scanf("%d", &a);
  for (int t = 1; t <= a; u[t] = {n, t}, t++) scanf("%d", &n);
  for (int t = 1; t <= 30; t++) {
    e = (1 << (t));
    sort(u + 1, u + a + 1, as);
    n = m = 0;
    x = y = 0;
    for (int i = 1; i <= a; i++) {
      if (u[i - 1].x / e != u[i].x / e) {
        n = m = 0;
      }
      if (u[i].x % e >= e / 2)
        y += n, m++;
      else
        x += m, n++;
    }
    if (x > y) {
      for (int t = 1; t <= a; t++) u[t].x ^= (e / 2);
      z += y, k += e / 2;
    } else
      z += x;
  }
  printf("%lld %d", z, k);
}

```