#include <bits/stdc++.h>
using namespace std;
pair<long long, long long> find(long long bit, vector<vector<long long>> a) {
  if (bit < 0) return {0, 0};
  long long cnt1 = 0, cnt2 = 0;
  for (long long i = 0; i < a.size(); i++) {
    long long c1 = 0, c2 = 0;
    for (auto y : a[i]) {
      long long c = (1 << bit) & y;
      if (c)
        c1++;
      else
        cnt1 += c1;
      if (c)
        cnt2 += c2;
      else
        c2++;
    }
  }
  long long inv_cnt = min(cnt1, cnt2);
  long long x = 0;
  if (cnt2 < cnt1) x = 1 << bit;
  vector<vector<long long>> v;
  for (auto i : a) {
    vector<long long> v1, v2;
    for (auto y : i) {
      long long c = (1 << bit) & y;
      if (c)
        v2.push_back(y);
      else
        v1.push_back(y);
    }
    if (v1.size()) v.push_back(v1);
    if (v2.size()) v.push_back(v2);
  }
  pair<long long, long long> ans = {0, 0};
  if (v.size()) ans = find(bit - 1, v);
  return {inv_cnt + ans.first, x + ans.second};
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  vector<vector<long long>> v;
  v.push_back(a);
  pair<long long, long long> ans = find(30, v);
  cout << ans.first << " " << ans.second;
}

### Prompt

Generate 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;
pair<long long, long long> find(long long bit, vector<vector<long long>> a) {
  if (bit < 0) return {0, 0};
  long long cnt1 = 0, cnt2 = 0;
  for (long long i = 0; i < a.size(); i++) {
    long long c1 = 0, c2 = 0;
    for (auto y : a[i]) {
      long long c = (1 << bit) & y;
      if (c)
        c1++;
      else
        cnt1 += c1;
      if (c)
        cnt2 += c2;
      else
        c2++;
    }
  }
  long long inv_cnt = min(cnt1, cnt2);
  long long x = 0;
  if (cnt2 < cnt1) x = 1 << bit;
  vector<vector<long long>> v;
  for (auto i : a) {
    vector<long long> v1, v2;
    for (auto y : i) {
      long long c = (1 << bit) & y;
      if (c)
        v2.push_back(y);
      else
        v1.push_back(y);
    }
    if (v1.size()) v.push_back(v1);
    if (v2.size()) v.push_back(v2);
  }
  pair<long long, long long> ans = {0, 0};
  if (v.size()) ans = find(bit - 1, v);
  return {inv_cnt + ans.first, x + ans.second};
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  vector<vector<long long>> v;
  v.push_back(a);
  pair<long long, long long> ans = find(30, v);
  cout << ans.first << " " << ans.second;
}

```