#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T &n) {
  T w = 1;
  n = 0;
  char ch = getchar();
  while (!isdigit(ch) && ch != EOF) {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (isdigit(ch) && ch != EOF) {
    n = (n << 1) + (n << 3) + (ch ^ 48);
    ch = getchar();
  }
  n *= w;
}
template <typename T>
inline void write(T x) {
  unsigned long long y = 0;
  T l = 0;
  if (x < 0) {
    x = -x;
    putchar('-');
  }
  if (!x) {
    putchar(48);
    return;
  }
  while (x) {
    y = y * 10 + x % 10;
    x /= 10;
    l++;
  }
  while (l) {
    putchar(y % 10 + 48);
    y /= 10;
    l--;
  }
}
template <typename T>
inline void writeln(T x) {
  write(x);
  puts("");
}
template <typename T>
inline void writes(T x) {
  write(x);
  putchar(' ');
}
template <typename T>
inline void checkmax(T &a, T b) {
  a = a > b ? a : b;
}
template <typename T>
inline void checkmin(T &a, T b) {
  a = a < b ? a : b;
}
int a[300100], c[300100], now[300100];
long long no;
inline void merge(int l, int r) {
  int mid = (l + r) >> 1;
  int i = l, j = mid + 1, p = l;
  for (; i <= mid && j <= r;)
    if (c[i] > c[j]) {
      no += mid - i + 1;
      now[p++] = c[j++];
    } else
      now[p++] = c[i++];
  while (i <= mid) now[p++] = c[i++];
  while (j <= r) now[p++] = c[j++];
  for (int i = (l); i <= (r); i++) c[i] = now[i];
}
inline void mergesort(int l, int r) {
  if (l >= r) return;
  int mid = (l + r) >> 1;
  mergesort(l, mid);
  mergesort(mid + 1, r);
  merge(l, r);
}
int main() {
  int n, ans = 0;
  read(n);
  for (int i = (1); i <= (n); i++) read(a[i]);
  for (int i = (30); i >= (0); i--) {
    int sum1 = 0, sum0 = 0;
    long long ni1 = 0, ni2 = 0;
    for (int j = (1); j <= (n); j++) c[j] = a[j];
    no = 0;
    mergesort(1, n);
    ni1 = no;
    for (int j = (1); j <= (n); j++) c[j] = a[j] ^ (1 << i);
    no = 0;
    mergesort(1, n);
    ni2 = no;
    if (ni1 > ni2) {
      for (int j = (1); j <= (n); j++) a[j] ^= (1 << i);
      ans += (1 << i);
    }
  }
  for (int i = (1); i <= (n); i++) c[i] = a[i];
  no = 0;
  mergesort(1, n);
  writes(no);
  writeln(ans);
  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;
template <typename T>
inline void read(T &n) {
  T w = 1;
  n = 0;
  char ch = getchar();
  while (!isdigit(ch) && ch != EOF) {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (isdigit(ch) && ch != EOF) {
    n = (n << 1) + (n << 3) + (ch ^ 48);
    ch = getchar();
  }
  n *= w;
}
template <typename T>
inline void write(T x) {
  unsigned long long y = 0;
  T l = 0;
  if (x < 0) {
    x = -x;
    putchar('-');
  }
  if (!x) {
    putchar(48);
    return;
  }
  while (x) {
    y = y * 10 + x % 10;
    x /= 10;
    l++;
  }
  while (l) {
    putchar(y % 10 + 48);
    y /= 10;
    l--;
  }
}
template <typename T>
inline void writeln(T x) {
  write(x);
  puts("");
}
template <typename T>
inline void writes(T x) {
  write(x);
  putchar(' ');
}
template <typename T>
inline void checkmax(T &a, T b) {
  a = a > b ? a : b;
}
template <typename T>
inline void checkmin(T &a, T b) {
  a = a < b ? a : b;
}
int a[300100], c[300100], now[300100];
long long no;
inline void merge(int l, int r) {
  int mid = (l + r) >> 1;
  int i = l, j = mid + 1, p = l;
  for (; i <= mid && j <= r;)
    if (c[i] > c[j]) {
      no += mid - i + 1;
      now[p++] = c[j++];
    } else
      now[p++] = c[i++];
  while (i <= mid) now[p++] = c[i++];
  while (j <= r) now[p++] = c[j++];
  for (int i = (l); i <= (r); i++) c[i] = now[i];
}
inline void mergesort(int l, int r) {
  if (l >= r) return;
  int mid = (l + r) >> 1;
  mergesort(l, mid);
  mergesort(mid + 1, r);
  merge(l, r);
}
int main() {
  int n, ans = 0;
  read(n);
  for (int i = (1); i <= (n); i++) read(a[i]);
  for (int i = (30); i >= (0); i--) {
    int sum1 = 0, sum0 = 0;
    long long ni1 = 0, ni2 = 0;
    for (int j = (1); j <= (n); j++) c[j] = a[j];
    no = 0;
    mergesort(1, n);
    ni1 = no;
    for (int j = (1); j <= (n); j++) c[j] = a[j] ^ (1 << i);
    no = 0;
    mergesort(1, n);
    ni2 = no;
    if (ni1 > ni2) {
      for (int j = (1); j <= (n); j++) a[j] ^= (1 << i);
      ans += (1 << i);
    }
  }
  for (int i = (1); i <= (n); i++) c[i] = a[i];
  no = 0;
  mergesort(1, n);
  writes(no);
  writeln(ans);
  return 0;
}

```