#include <bits/stdc++.h>
using namespace std;
const int max_n = 3e5 + 10, inf = 1000111222;
struct node {
  int p0, p1;
  int cnt;
  node() {
    p0 = p1 = 0;
    cnt = 0;
  }
};
int root = 1;
node A[max_n * 33];
int last_used = 1;
int new_node() { return ++last_used; }
int a[max_n];
long long cost_if_set[33][2];
void add_number(int a) {
  int cur = root;
  for (int i = 30; i >= 0; i--) {
    if (a & (1ll << i)) {
      if (A[cur].p0) {
        cost_if_set[i][0] += A[A[cur].p0].cnt;
      }
      if (!A[cur].p1) {
        A[cur].p1 = new_node();
      }
      cur = A[cur].p1;
      A[cur].cnt++;
    } else {
      if (A[cur].p1) {
        cost_if_set[i][1] += A[A[cur].p1].cnt;
      }
      if (!A[cur].p0) {
        A[cur].p0 = new_node();
      }
      cur = A[cur].p0;
      A[cur].cnt++;
    }
  }
}
int n;
void calc_costs() {
  for (int i = n; i >= 1; i--) {
    add_number(a[i]);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  calc_costs();
  long long ans = 0;
  int x = 0;
  for (int i = 0; i < 31; i++) {
    if (cost_if_set[i][0] <= cost_if_set[i][1]) {
      ans += cost_if_set[i][0];
    } else {
      x |= (1ll << i);
      ans += cost_if_set[i][1];
    }
  }
  cout << ans << " " << x << "\n";
}

### 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;
const int max_n = 3e5 + 10, inf = 1000111222;
struct node {
  int p0, p1;
  int cnt;
  node() {
    p0 = p1 = 0;
    cnt = 0;
  }
};
int root = 1;
node A[max_n * 33];
int last_used = 1;
int new_node() { return ++last_used; }
int a[max_n];
long long cost_if_set[33][2];
void add_number(int a) {
  int cur = root;
  for (int i = 30; i >= 0; i--) {
    if (a & (1ll << i)) {
      if (A[cur].p0) {
        cost_if_set[i][0] += A[A[cur].p0].cnt;
      }
      if (!A[cur].p1) {
        A[cur].p1 = new_node();
      }
      cur = A[cur].p1;
      A[cur].cnt++;
    } else {
      if (A[cur].p1) {
        cost_if_set[i][1] += A[A[cur].p1].cnt;
      }
      if (!A[cur].p0) {
        A[cur].p0 = new_node();
      }
      cur = A[cur].p0;
      A[cur].cnt++;
    }
  }
}
int n;
void calc_costs() {
  for (int i = n; i >= 1; i--) {
    add_number(a[i]);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  calc_costs();
  long long ans = 0;
  int x = 0;
  for (int i = 0; i < 31; i++) {
    if (cost_if_set[i][0] <= cost_if_set[i][1]) {
      ans += cost_if_set[i][0];
    } else {
      x |= (1ll << i);
      ans += cost_if_set[i][1];
    }
  }
  cout << ans << " " << x << "\n";
}

```