#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <class T>
using pque = priority_queue<T, vector<T>, greater<T>>;
constexpr int inf = 1000000010;
constexpr long long INF = 1000000000000000010;
constexpr int mod1e9 = 1000000007;
constexpr int mod998 = 998244353;
constexpr long double eps = 1e-12;
constexpr long double pi = 3.141592653589793238;
constexpr long long ten(int n) { return n ? 10 * ten(n - 1) : 1; };
int dx[] = {1, 0, -1, 0, 1, 1, -1, -1};
int dy[] = {0, 1, 0, -1, 1, -1, 1, -1};
void fail() {
  cout << "-1\n";
  exit(0);
}
void no() {
  cout << "No\n";
  exit(0);
}
template <class T, class U>
inline bool chmax(T &a, const U &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T, class U>
inline bool chmin(T &a, const U &b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
istream &operator>>(istream &s, vector<T> &v) {
  for (auto &e : v) s >> e;
  return s;
}
template <class T>
ostream &operator<<(ostream &s, const vector<T> &v) {
  for (auto &e : v) s << e << ' ';
  return s;
}
struct fastio {
  fastio() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(20);
    cerr << fixed << setprecision(20);
  }
} fastio_;
int main() {
  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);
  long long inv = 0;
  int ans = 0;
  for (int i = 29; i >= 0; i--) {
    vector<vector<long long>> nxt;
    long long sum = 0;
    long long c = 0, r = 0;
    for (auto vec : v) {
      long long zero = 0, one = 0, cinv = 0;
      vector<long long> x, y;
      for (int num : vec) {
        if (num >> i & 1) {
          one++;
          x.push_back(num);
        } else {
          zero++;
          cinv += one;
          y.push_back(num);
        }
      }
      long long rinv = zero * one - cinv;
      c += cinv;
      r += rinv;
      if (x.size()) nxt.push_back(x);
      if (y.size()) nxt.push_back(y);
    }
    if (c > r) {
      inv += r;
      ans ^= 1 << i;
    } else
      inv += c;
    v = nxt;
  }
  cout << inv << ' ' << ans << '\n';
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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")
#pragma GCC optimize("unroll-loops")
using namespace std;
template <class T>
using pque = priority_queue<T, vector<T>, greater<T>>;
constexpr int inf = 1000000010;
constexpr long long INF = 1000000000000000010;
constexpr int mod1e9 = 1000000007;
constexpr int mod998 = 998244353;
constexpr long double eps = 1e-12;
constexpr long double pi = 3.141592653589793238;
constexpr long long ten(int n) { return n ? 10 * ten(n - 1) : 1; };
int dx[] = {1, 0, -1, 0, 1, 1, -1, -1};
int dy[] = {0, 1, 0, -1, 1, -1, 1, -1};
void fail() {
  cout << "-1\n";
  exit(0);
}
void no() {
  cout << "No\n";
  exit(0);
}
template <class T, class U>
inline bool chmax(T &a, const U &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T, class U>
inline bool chmin(T &a, const U &b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
istream &operator>>(istream &s, vector<T> &v) {
  for (auto &e : v) s >> e;
  return s;
}
template <class T>
ostream &operator<<(ostream &s, const vector<T> &v) {
  for (auto &e : v) s << e << ' ';
  return s;
}
struct fastio {
  fastio() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(20);
    cerr << fixed << setprecision(20);
  }
} fastio_;
int main() {
  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);
  long long inv = 0;
  int ans = 0;
  for (int i = 29; i >= 0; i--) {
    vector<vector<long long>> nxt;
    long long sum = 0;
    long long c = 0, r = 0;
    for (auto vec : v) {
      long long zero = 0, one = 0, cinv = 0;
      vector<long long> x, y;
      for (int num : vec) {
        if (num >> i & 1) {
          one++;
          x.push_back(num);
        } else {
          zero++;
          cinv += one;
          y.push_back(num);
        }
      }
      long long rinv = zero * one - cinv;
      c += cinv;
      r += rinv;
      if (x.size()) nxt.push_back(x);
      if (y.size()) nxt.push_back(y);
    }
    if (c > r) {
      inv += r;
      ans ^= 1 << i;
    } else
      inv += c;
    v = nxt;
  }
  cout << inv << ' ' << ans << '\n';
}

```