#include <bits/stdc++.h>
using namespace std;
template <class C>
void min_self(C &a, C b) {
  a = min(a, b);
}
template <class C>
void max_self(C &a, C b) {
  a = max(a, b);
}
const long long MOD = 1000000007;
long long mod(long long n, long long m = MOD) {
  n %= m, n += m, n %= m;
  return n;
}
const int MAXN = 1e5 + 5;
const int LOGN = 21;
const long long INF = 1e14;
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
template <class T1, class T2>
void add(T1 &x, T2 y, long long m = MOD) {
  x += y;
  if (x >= m) x -= m;
}
template <class T1, class T2>
void sub(T1 &x, T2 y, long long m = MOD) {
  x -= y;
  if (x < 0) x += m;
}
vector<vector<int>> groups[32];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  vector<int> temp;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    temp.push_back(i);
  }
  groups[31].push_back(temp);
  long long ans = 0, inversions = 0;
  for (long long bit = 31; bit >= 0; bit--) {
    long long pairs01 = 0, pairs10 = 0;
    for (auto v : groups[bit]) {
      int sz = v.size();
      vector<int> cnt(2, 0);
      vector<int> ones, zeros;
      for (int i = 0; i < sz; i++) {
        if ((a[v[i]] >> bit) & 1) {
          pairs01 += cnt[0];
          cnt[1]++;
          ones.push_back(v[i]);
        } else {
          pairs10 += cnt[1];
          cnt[0]++;
          zeros.push_back(v[i]);
        }
      }
      if (bit > 0) {
        if (!ones.empty()) groups[bit - 1].push_back(ones);
        if (!zeros.empty()) groups[bit - 1].push_back(zeros);
      }
    }
    if (pairs10 > pairs01) {
      ans |= (1 << bit);
    }
    inversions += min(pairs10, pairs01);
  }
  cout << inversions << " " << ans, cout << "\n";
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}

### Prompt

Your task is to create 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;
template <class C>
void min_self(C &a, C b) {
  a = min(a, b);
}
template <class C>
void max_self(C &a, C b) {
  a = max(a, b);
}
const long long MOD = 1000000007;
long long mod(long long n, long long m = MOD) {
  n %= m, n += m, n %= m;
  return n;
}
const int MAXN = 1e5 + 5;
const int LOGN = 21;
const long long INF = 1e14;
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
template <class T1, class T2>
void add(T1 &x, T2 y, long long m = MOD) {
  x += y;
  if (x >= m) x -= m;
}
template <class T1, class T2>
void sub(T1 &x, T2 y, long long m = MOD) {
  x -= y;
  if (x < 0) x += m;
}
vector<vector<int>> groups[32];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  vector<int> temp;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    temp.push_back(i);
  }
  groups[31].push_back(temp);
  long long ans = 0, inversions = 0;
  for (long long bit = 31; bit >= 0; bit--) {
    long long pairs01 = 0, pairs10 = 0;
    for (auto v : groups[bit]) {
      int sz = v.size();
      vector<int> cnt(2, 0);
      vector<int> ones, zeros;
      for (int i = 0; i < sz; i++) {
        if ((a[v[i]] >> bit) & 1) {
          pairs01 += cnt[0];
          cnt[1]++;
          ones.push_back(v[i]);
        } else {
          pairs10 += cnt[1];
          cnt[0]++;
          zeros.push_back(v[i]);
        }
      }
      if (bit > 0) {
        if (!ones.empty()) groups[bit - 1].push_back(ones);
        if (!zeros.empty()) groups[bit - 1].push_back(zeros);
      }
    }
    if (pairs10 > pairs01) {
      ans |= (1 << bit);
    }
    inversions += min(pairs10, pairs01);
  }
  cout << inversions << " " << ans, cout << "\n";
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}

```