#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5;
const int C = 30 + 5;
const int M = N * 30 + 5;
const int INF = 0x3f3f3f3f;
const long long INF_LL = 0x3f3f3f3f3f3f3f3f;
int ch[M][2];
vector<int> pos[M];
int nidx = 0;
int rt = 0;
void Insert(int &o, int dep, int x, int _pos) {
  if (o == 0) o = ++nidx;
  if (dep < 0) {
    pos[o].push_back(_pos);
    return;
  }
  int op = (x >> dep) & 1;
  Insert(ch[o][op], dep - 1, x, _pos);
}
long long f[C][2];
void DFS(int o, int dep, vector<int> &vec) {
  if (pos[o].size()) {
    vec = pos[o];
  } else {
    vector<int> a, b;
    if (ch[o][0]) DFS(ch[o][0], dep - 1, a);
    if (ch[o][1]) DFS(ch[o][1], dep - 1, b);
    for (auto x : b)
      f[dep][0] += lower_bound(a.begin(), a.end(), x) - a.begin();
    for (auto x : a)
      f[dep][1] += lower_bound(b.begin(), b.end(), x) - b.begin();
    vec.resize(a.size() + b.size());
    merge(a.begin(), a.end(), b.begin(), b.end(), vec.begin());
  }
}
int a[N];
long long sumInv = 0;
int main() {
  ios::sync_with_stdio(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    Insert(rt, 30, a[i], i);
  }
  vector<int> vec;
  DFS(rt, 30, vec);
  int ans = 0;
  for (int i = 30; i >= 0; i--) {
    sumInv += f[i][1];
    if (f[i][0] - f[i][1] < 0) {
      ans |= 1 << i;
      sumInv += f[i][0] - f[i][1];
    }
  }
  cout << sumInv << ' ' << ans << endl;
  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 N = 3e5 + 5;
const int C = 30 + 5;
const int M = N * 30 + 5;
const int INF = 0x3f3f3f3f;
const long long INF_LL = 0x3f3f3f3f3f3f3f3f;
int ch[M][2];
vector<int> pos[M];
int nidx = 0;
int rt = 0;
void Insert(int &o, int dep, int x, int _pos) {
  if (o == 0) o = ++nidx;
  if (dep < 0) {
    pos[o].push_back(_pos);
    return;
  }
  int op = (x >> dep) & 1;
  Insert(ch[o][op], dep - 1, x, _pos);
}
long long f[C][2];
void DFS(int o, int dep, vector<int> &vec) {
  if (pos[o].size()) {
    vec = pos[o];
  } else {
    vector<int> a, b;
    if (ch[o][0]) DFS(ch[o][0], dep - 1, a);
    if (ch[o][1]) DFS(ch[o][1], dep - 1, b);
    for (auto x : b)
      f[dep][0] += lower_bound(a.begin(), a.end(), x) - a.begin();
    for (auto x : a)
      f[dep][1] += lower_bound(b.begin(), b.end(), x) - b.begin();
    vec.resize(a.size() + b.size());
    merge(a.begin(), a.end(), b.begin(), b.end(), vec.begin());
  }
}
int a[N];
long long sumInv = 0;
int main() {
  ios::sync_with_stdio(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    Insert(rt, 30, a[i], i);
  }
  vector<int> vec;
  DFS(rt, 30, vec);
  int ans = 0;
  for (int i = 30; i >= 0; i--) {
    sumInv += f[i][1];
    if (f[i][0] - f[i][1] < 0) {
      ans |= 1 << i;
      sumInv += f[i][0] - f[i][1];
    }
  }
  cout << sumInv << ' ' << ans << endl;
  return 0;
}

```