#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast")
#pragma GCC target( \
    "fma,sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
#pragma GCC optimize("unroll-loops")
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << '{';
  string sep;
  for (const auto &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
template <typename T, size_t size>
ostream &operator<<(ostream &os, const array<T, size> &arr) {
  os << '{';
  string sep;
  for (const auto &x : arr) os << sep << x, sep = ", ";
  return os << '}';
}
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << ' ' << H;
  debug_out(T...);
}
vector<int> a;
const int MAX_N = 3e5 + 5;
vector<int> add(MAX_N);
long long solve(int l, int r) {
  if (l == r) {
    return 0;
  }
  long long res = 0;
  const int mid = l + (r - l) / 2;
  res += solve(l, mid);
  res += solve(mid + 1, r);
  int l1 = l;
  int l2 = mid + 1;
  int p = 0;
  while (l1 <= mid || l2 <= r) {
    if (l1 > mid) {
      add[p] = a[l2];
      ++l2;
    } else if (l2 > r) {
      add[p] = a[l1];
      ++l1;
    } else {
      if (a[l1] <= a[l2]) {
        add[p] = a[l1];
        ++l1;
      } else {
        add[p] = a[l2];
        ++l2;
        res += (mid - l1 + 1);
      }
    }
    ++p;
  }
  for (int i = l; i <= r; i++) {
    a[i] = add[i - l];
  }
  return res;
}
long long inversions(const vector<int> &arr) {
  a = arr;
  const int t = a.size();
  return solve(0, t - 1);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> arr(n);
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  int x = 0;
  long long before = inversions(arr);
  for (int i = 0; i < 31; i++) {
    for (int j = 0; j < n; j++) {
      arr[j] ^= (1 << i);
    }
    long long after = inversions(arr);
    if (after < before) {
      before = after;
      x |= (1 << i);
    } else {
      for (int j = 0; j < n; j++) {
        arr[j] ^= (1 << i);
      }
    }
  }
  cout << inversions(arr) << ' ' << x << '\n';
  return 0;
}

### Prompt

Please create a solution in cpp 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;
#pragma GCC optimize("Ofast")
#pragma GCC target( \
    "fma,sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
#pragma GCC optimize("unroll-loops")
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << '{';
  string sep;
  for (const auto &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
template <typename T, size_t size>
ostream &operator<<(ostream &os, const array<T, size> &arr) {
  os << '{';
  string sep;
  for (const auto &x : arr) os << sep << x, sep = ", ";
  return os << '}';
}
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << ' ' << H;
  debug_out(T...);
}
vector<int> a;
const int MAX_N = 3e5 + 5;
vector<int> add(MAX_N);
long long solve(int l, int r) {
  if (l == r) {
    return 0;
  }
  long long res = 0;
  const int mid = l + (r - l) / 2;
  res += solve(l, mid);
  res += solve(mid + 1, r);
  int l1 = l;
  int l2 = mid + 1;
  int p = 0;
  while (l1 <= mid || l2 <= r) {
    if (l1 > mid) {
      add[p] = a[l2];
      ++l2;
    } else if (l2 > r) {
      add[p] = a[l1];
      ++l1;
    } else {
      if (a[l1] <= a[l2]) {
        add[p] = a[l1];
        ++l1;
      } else {
        add[p] = a[l2];
        ++l2;
        res += (mid - l1 + 1);
      }
    }
    ++p;
  }
  for (int i = l; i <= r; i++) {
    a[i] = add[i - l];
  }
  return res;
}
long long inversions(const vector<int> &arr) {
  a = arr;
  const int t = a.size();
  return solve(0, t - 1);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> arr(n);
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  int x = 0;
  long long before = inversions(arr);
  for (int i = 0; i < 31; i++) {
    for (int j = 0; j < n; j++) {
      arr[j] ^= (1 << i);
    }
    long long after = inversions(arr);
    if (after < before) {
      before = after;
      x |= (1 << i);
    } else {
      for (int j = 0; j < n; j++) {
        arr[j] ^= (1 << i);
      }
    }
  }
  cout << inversions(arr) << ' ' << x << '\n';
  return 0;
}

```