#include <bits/stdc++.h>
using namespace std;
template <typename Tp>
inline void read(Tp &x) {
  static char c;
  static bool neg;
  x = 0, c = getchar(), neg = false;
  for (; !isdigit(c); c = getchar()) {
    if (c == '-') {
      neg = true;
    }
  }
  for (; isdigit(c); c = getchar()) {
    x = x * 10 + c - '0';
  }
  if (neg) {
    x = -x;
  }
}
const int N = 3e5 + 5;
int n;
int a[N];
struct Set {
  vector<int> fir;
  vector<int> sec;
  inline void push(int len, int x) {
    for (int i = 1, t = (1 << x); i <= len; ++i) {
      if (a[i] & t) {
        fir.emplace_back(i);
      } else {
        sec.emplace_back(i);
      }
    }
  }
  inline void push(const vector<int> &vec, int x) {
    int t = (1 << x);
    for (const auto &v : vec) {
      if (a[v] & t) {
        fir.emplace_back(v);
      } else {
        sec.emplace_back(v);
      }
    }
  }
};
const int MX = 30;
vector<Set> vec[MX];
int main() {
  read(n);
  for (int i = 1; i <= n; ++i) {
    read(a[i]);
  }
  vec[MX - 1].emplace_back(Set());
  vec[MX - 1].back().push(n, MX - 1);
  long long x = 0, cnt = 0, ans0, ans1;
  for (int i = MX - 1; i >= 0; --i) {
    ans0 = ans1 = 0LL;
    for (const auto &x : vec[i]) {
      auto fir = &x.fir, sec = &x.sec;
      size_t pos = 0UL;
      for (size_t now = 0UL; now < fir->size(); ++now) {
        while (pos < sec->size() &&
               sec->operator[](pos) <= fir->operator[](now)) {
          ++pos;
        }
        ans0 += sec->size() - pos;
        ans1 += pos;
      }
      if (i > 0) {
        if (!fir->empty()) {
          vec[i - 1].emplace_back(Set());
          vec[i - 1].back().push(*fir, i - 1);
        }
        if (!sec->empty()) {
          vec[i - 1].emplace_back(Set());
          vec[i - 1].back().push(*sec, i - 1);
        }
      }
    }
    if (ans0 > ans1) {
      x |= (1 << i);
      cnt += ans1;
    } else {
      cnt += ans0;
    }
  }
  printf("%lld %lld\n", cnt, x);
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 <typename Tp>
inline void read(Tp &x) {
  static char c;
  static bool neg;
  x = 0, c = getchar(), neg = false;
  for (; !isdigit(c); c = getchar()) {
    if (c == '-') {
      neg = true;
    }
  }
  for (; isdigit(c); c = getchar()) {
    x = x * 10 + c - '0';
  }
  if (neg) {
    x = -x;
  }
}
const int N = 3e5 + 5;
int n;
int a[N];
struct Set {
  vector<int> fir;
  vector<int> sec;
  inline void push(int len, int x) {
    for (int i = 1, t = (1 << x); i <= len; ++i) {
      if (a[i] & t) {
        fir.emplace_back(i);
      } else {
        sec.emplace_back(i);
      }
    }
  }
  inline void push(const vector<int> &vec, int x) {
    int t = (1 << x);
    for (const auto &v : vec) {
      if (a[v] & t) {
        fir.emplace_back(v);
      } else {
        sec.emplace_back(v);
      }
    }
  }
};
const int MX = 30;
vector<Set> vec[MX];
int main() {
  read(n);
  for (int i = 1; i <= n; ++i) {
    read(a[i]);
  }
  vec[MX - 1].emplace_back(Set());
  vec[MX - 1].back().push(n, MX - 1);
  long long x = 0, cnt = 0, ans0, ans1;
  for (int i = MX - 1; i >= 0; --i) {
    ans0 = ans1 = 0LL;
    for (const auto &x : vec[i]) {
      auto fir = &x.fir, sec = &x.sec;
      size_t pos = 0UL;
      for (size_t now = 0UL; now < fir->size(); ++now) {
        while (pos < sec->size() &&
               sec->operator[](pos) <= fir->operator[](now)) {
          ++pos;
        }
        ans0 += sec->size() - pos;
        ans1 += pos;
      }
      if (i > 0) {
        if (!fir->empty()) {
          vec[i - 1].emplace_back(Set());
          vec[i - 1].back().push(*fir, i - 1);
        }
        if (!sec->empty()) {
          vec[i - 1].emplace_back(Set());
          vec[i - 1].back().push(*sec, i - 1);
        }
      }
    }
    if (ans0 > ans1) {
      x |= (1 << i);
      cnt += ans1;
    } else {
      cnt += ans0;
    }
  }
  printf("%lld %lld\n", cnt, x);
}

```