#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
const int maxn = 3e5 + 10;
const int inf = 0x3f3f3f3f;
const double eps = 1e-6;
int n, a[maxn];
long long mn[35][2];
struct node {
  vector<int> v;
  int d;
};
queue<node> q;
void bfs() {
  node c;
  c.d = 29;
  for (int i = 1; i <= n; i++) c.v.push_back(a[i]);
  q.push(c);
  while (!q.empty()) {
    node d = q.front();
    q.pop();
    node nxt1, nxt2;
    nxt1.d = d.d - 1;
    nxt2.d = d.d - 1;
    int num = 0;
    long long inv0 = 0, inv1 = 0;
    for (auto u : d.v) {
      if ((u & (1 << d.d)) > 0)
        num++;
      else
        inv0 += num;
    }
    num = 0;
    for (auto u : d.v) {
      if ((u & (1 << d.d)) == 0)
        num++;
      else
        inv1 += num;
    }
    mn[d.d][0] += inv0;
    mn[d.d][1] += inv1;
    if (d.d == 0) continue;
    for (auto u : d.v) {
      if ((u & (1 << d.d)) > 0)
        nxt1.v.push_back(u);
      else
        nxt2.v.push_back(u);
    }
    if ((int)nxt1.v.size() > 0) q.push(nxt1);
    if ((int)nxt2.v.size() > 0) q.push(nxt2);
  }
}
int main() {
  std::ios::sync_with_stdio(false), cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  bfs();
  long long res = 0, x = 0;
  for (int i = 0; i <= 29; i++)
    if (mn[i][1] < mn[i][0])
      res += mn[i][1], x |= (1 << i);
    else
      res += mn[i][0];
  cout << res << ' ' << x << '\n';
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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;
const int mod = 998244353;
const int maxn = 3e5 + 10;
const int inf = 0x3f3f3f3f;
const double eps = 1e-6;
int n, a[maxn];
long long mn[35][2];
struct node {
  vector<int> v;
  int d;
};
queue<node> q;
void bfs() {
  node c;
  c.d = 29;
  for (int i = 1; i <= n; i++) c.v.push_back(a[i]);
  q.push(c);
  while (!q.empty()) {
    node d = q.front();
    q.pop();
    node nxt1, nxt2;
    nxt1.d = d.d - 1;
    nxt2.d = d.d - 1;
    int num = 0;
    long long inv0 = 0, inv1 = 0;
    for (auto u : d.v) {
      if ((u & (1 << d.d)) > 0)
        num++;
      else
        inv0 += num;
    }
    num = 0;
    for (auto u : d.v) {
      if ((u & (1 << d.d)) == 0)
        num++;
      else
        inv1 += num;
    }
    mn[d.d][0] += inv0;
    mn[d.d][1] += inv1;
    if (d.d == 0) continue;
    for (auto u : d.v) {
      if ((u & (1 << d.d)) > 0)
        nxt1.v.push_back(u);
      else
        nxt2.v.push_back(u);
    }
    if ((int)nxt1.v.size() > 0) q.push(nxt1);
    if ((int)nxt2.v.size() > 0) q.push(nxt2);
  }
}
int main() {
  std::ios::sync_with_stdio(false), cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  bfs();
  long long res = 0, x = 0;
  for (int i = 0; i <= 29; i++)
    if (mn[i][1] < mn[i][0])
      res += mn[i][1], x |= (1 << i);
    else
      res += mn[i][0];
  cout << res << ' ' << x << '\n';
  return 0;
}

```