#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
using ll = long long;
using pii = pair<int, int>;
const int MOD = 1e9 + 7;
const int INF = 1e9 + 1e2;
void fin() {}
const bool flag = 0;
const int N = 3e5 + 10;
int a[N];
ll cnt[2][30];
void calc(int cur, vector<int> p) {
  if (cur == -1 || p.empty()) {
    return;
  }
  int z = 0, o = 0;
  vector<int> l, r;
  for (int i : p) {
    if (a[i] >> cur & 1) {
      l.push_back(i), o++, cnt[1][cur] += z;
    } else {
      r.push_back(i), z++, cnt[0][cur] += o;
    }
  }
  calc(cur - 1, l), calc(cur - 1, r);
}
void ma1n() {
  int n;
  cin >> n;
  vector<int> p;
  for (int i = 0; i < n; ++i) {
    cin >> a[i], p.push_back(i);
  }
  calc(29, p);
  ll ans = 0, first = 0;
  for (int i = 0; i < 30; ++i) {
    if (cnt[1][i] < cnt[0][i]) {
      ans += cnt[1][i], first |= (1 << i);
    } else {
      ans += cnt[0][i];
    }
  }
  cout << ans << " " << first;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr), fin();
  int ts = 1;
  if (flag) {
    cin >> ts;
  }
  while (ts--) {
    ma1n();
  }
  return 0;
}

### 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>
#pragma GCC optimize("O3")
using namespace std;
using ll = long long;
using pii = pair<int, int>;
const int MOD = 1e9 + 7;
const int INF = 1e9 + 1e2;
void fin() {}
const bool flag = 0;
const int N = 3e5 + 10;
int a[N];
ll cnt[2][30];
void calc(int cur, vector<int> p) {
  if (cur == -1 || p.empty()) {
    return;
  }
  int z = 0, o = 0;
  vector<int> l, r;
  for (int i : p) {
    if (a[i] >> cur & 1) {
      l.push_back(i), o++, cnt[1][cur] += z;
    } else {
      r.push_back(i), z++, cnt[0][cur] += o;
    }
  }
  calc(cur - 1, l), calc(cur - 1, r);
}
void ma1n() {
  int n;
  cin >> n;
  vector<int> p;
  for (int i = 0; i < n; ++i) {
    cin >> a[i], p.push_back(i);
  }
  calc(29, p);
  ll ans = 0, first = 0;
  for (int i = 0; i < 30; ++i) {
    if (cnt[1][i] < cnt[0][i]) {
      ans += cnt[1][i], first |= (1 << i);
    } else {
      ans += cnt[0][i];
    }
  }
  cout << ans << " " << first;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr), fin();
  int ts = 1;
  if (flag) {
    cin >> ts;
  }
  while (ts--) {
    ma1n();
  }
  return 0;
}

```