#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
template <class T>
inline void amin(T &x, const T &y) {
  if (y < x) x = y;
}
template <class T>
inline void amax(T &x, const T &y) {
  if (x < y) x = y;
}
template <class Iter>
void rprintf(const char *fmt, Iter begin, Iter end) {
  for (bool sp = 0; begin != end; ++begin) {
    if (sp)
      putchar(' ');
    else
      sp = true;
    printf(fmt, *begin);
  }
  putchar('\n');
}
int N;
int A[300011];
pair<int, int> P[300011];
vector<int> idx;
void MAIN() {
  scanf("%d", &N);
  for (int i = 0, i_len = (N); i < i_len; ++i) scanf("%d", A + i);
  for (int i = 0, i_len = (N); i < i_len; ++i) P[i] = make_pair(A[i], i);
  sort(P, P + N);
  int ans = 0;
  long long inv = 0;
  for (int t = 30; t--;) {
    int t1 = t + 1;
    long long one_zero = 0;
    long long zero_one = 0;
    for (int i = 0; i < N;) {
      int j = i;
      while (j < N && (P[i].first >> t1) == (P[j].first >> t1)) j++;
      idx.clear();
      for (int x = i; x < j; x++)
        if (~P[x].first >> t & 1) {
          idx.push_back(P[x].second);
        }
      sort(idx.begin(), idx.end());
      for (int x = i; x < j; x++)
        if (P[x].first >> t & 1) {
          int pos =
              lower_bound(idx.begin(), idx.end(), P[x].second) - idx.begin();
          zero_one += pos;
          one_zero += (int)idx.size() - pos;
        }
      i = j;
    }
    if (one_zero > zero_one) {
      ans |= 1 << t;
      inv += zero_one;
    } else {
      inv += one_zero;
    }
  }
  printf("%lld %d\n", inv, ans);
}
int main() {
  int TC = 1;
  for (int tc = 0, tc_len = (TC); tc < tc_len; ++tc) MAIN();
  return 0;
}

### Prompt

Please formulate 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>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
template <class T>
inline void amin(T &x, const T &y) {
  if (y < x) x = y;
}
template <class T>
inline void amax(T &x, const T &y) {
  if (x < y) x = y;
}
template <class Iter>
void rprintf(const char *fmt, Iter begin, Iter end) {
  for (bool sp = 0; begin != end; ++begin) {
    if (sp)
      putchar(' ');
    else
      sp = true;
    printf(fmt, *begin);
  }
  putchar('\n');
}
int N;
int A[300011];
pair<int, int> P[300011];
vector<int> idx;
void MAIN() {
  scanf("%d", &N);
  for (int i = 0, i_len = (N); i < i_len; ++i) scanf("%d", A + i);
  for (int i = 0, i_len = (N); i < i_len; ++i) P[i] = make_pair(A[i], i);
  sort(P, P + N);
  int ans = 0;
  long long inv = 0;
  for (int t = 30; t--;) {
    int t1 = t + 1;
    long long one_zero = 0;
    long long zero_one = 0;
    for (int i = 0; i < N;) {
      int j = i;
      while (j < N && (P[i].first >> t1) == (P[j].first >> t1)) j++;
      idx.clear();
      for (int x = i; x < j; x++)
        if (~P[x].first >> t & 1) {
          idx.push_back(P[x].second);
        }
      sort(idx.begin(), idx.end());
      for (int x = i; x < j; x++)
        if (P[x].first >> t & 1) {
          int pos =
              lower_bound(idx.begin(), idx.end(), P[x].second) - idx.begin();
          zero_one += pos;
          one_zero += (int)idx.size() - pos;
        }
      i = j;
    }
    if (one_zero > zero_one) {
      ans |= 1 << t;
      inv += zero_one;
    } else {
      inv += one_zero;
    }
  }
  printf("%lld %d\n", inv, ans);
}
int main() {
  int TC = 1;
  for (int tc = 0, tc_len = (TC); tc < tc_len; ++tc) MAIN();
  return 0;
}

```