#include <bits/stdc++.h>
using namespace std;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
vector<long long> inversions(30), antinversions(30);
vector<vector<long long>> bits;
void solve(vector<long long>& indexes, long long bit) {
  if (bit < 0) return;
  vector<long long> ones, zeroes;
  long long cnt1 = 0, cnt0 = 0;
  for (long long i = 0; i < indexes.size(); i++) {
    if (bits[indexes[i]][bit]) {
      cnt1++;
      antinversions[bit] += cnt0;
      ones.push_back(indexes[i]);
    } else {
      cnt0++;
      inversions[bit] += cnt1;
      zeroes.push_back(indexes[i]);
    }
  }
  if (ones.size() > 0) {
    solve(ones, bit - 1);
  }
  if (zeroes.size() > 0) {
    solve(zeroes, bit - 1);
  }
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n;
  cin >> n;
  vector<long long> a(n);
  bits.resize(n, vector<long long>(30));
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    for (long long j = 29; j >= 0; j--) {
      if (a[i] & (1 << j)) {
        bits[i][j] = 1;
      }
    }
  }
  vector<long long> indexes(n);
  for (long long i = 0; i < n; i++) {
    indexes[i] = i;
  }
  solve(indexes, 29);
  long long x = 0;
  long long inv = 0;
  for (long long i = 29; i >= 0; i--) {
    if (inversions[i] > antinversions[i]) {
      x |= (1 << i);
      inv += antinversions[i];
    } else {
      inv += inversions[i];
    }
  }
  cout << inv << " " << x << endl;
}

### Prompt

In CPP, your task is to solve 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>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
vector<long long> inversions(30), antinversions(30);
vector<vector<long long>> bits;
void solve(vector<long long>& indexes, long long bit) {
  if (bit < 0) return;
  vector<long long> ones, zeroes;
  long long cnt1 = 0, cnt0 = 0;
  for (long long i = 0; i < indexes.size(); i++) {
    if (bits[indexes[i]][bit]) {
      cnt1++;
      antinversions[bit] += cnt0;
      ones.push_back(indexes[i]);
    } else {
      cnt0++;
      inversions[bit] += cnt1;
      zeroes.push_back(indexes[i]);
    }
  }
  if (ones.size() > 0) {
    solve(ones, bit - 1);
  }
  if (zeroes.size() > 0) {
    solve(zeroes, bit - 1);
  }
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n;
  cin >> n;
  vector<long long> a(n);
  bits.resize(n, vector<long long>(30));
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    for (long long j = 29; j >= 0; j--) {
      if (a[i] & (1 << j)) {
        bits[i][j] = 1;
      }
    }
  }
  vector<long long> indexes(n);
  for (long long i = 0; i < n; i++) {
    indexes[i] = i;
  }
  solve(indexes, 29);
  long long x = 0;
  long long inv = 0;
  for (long long i = 29; i >= 0; i--) {
    if (inversions[i] > antinversions[i]) {
      x |= (1 << i);
      inv += antinversions[i];
    } else {
      inv += inversions[i];
    }
  }
  cout << inv << " " << x << endl;
}

```