#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
  if (f < 0) x = -x;
}
template <typename T>
inline void write(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
int n;
int a[300010];
int num = 0;
int ans = 0;
int stk[300010][35];
long long suml[35][2];
void work(int len, int now) {
  if (len <= 0) return;
  long long gs0 = 0, gs1 = 0, a1 = 0, a2 = 0;
  int top = 0;
  if (now >= 1) top = 0;
  for (int i = 1; i <= len; i++) {
    if ((a[stk[i][now]] >> now) & 1) {
      a1 += gs0;
      ++gs1;
      if (now >= 1) top++, stk[top][now - 1] = stk[i][now];
    } else {
      a2 += gs1;
      ++gs0;
    }
  }
  suml[now][1] += (long long)a1;
  suml[now][0] += (long long)a2;
  if (now == 0) return;
  work(top, now - 1);
  top = 0;
  for (int i = 1; i <= len; i++) {
    if (!((a[stk[i][now]] >> now) & 1)) top++, stk[top][now - 1] = stk[i][now];
  }
  work(top, now - 1);
  return;
}
int main() {
  read(n);
  int x = 0;
  for (int i = 1; i <= n; i++) {
    read(a[i]);
    x = max(x, a[i]);
  }
  int loc = 30;
  while (!((1 << loc) & x) && loc >= 1) loc--;
  int top = 0;
  for (int i = 1; i <= n; i++) {
    top++;
    stk[top][loc] = i;
  }
  work(top, loc);
  long long ans2 = 0ll;
  for (int i = 0; i <= loc; i++) {
    if (suml[i][0] <= suml[i][1]) {
      ans2 += suml[i][0];
    } else {
      ans2 += suml[i][1];
      ans |= (1 << i);
    }
  }
  write(ans2);
  putchar(' ');
  write(ans);
  putchar('\n');
  return 0;
}

### 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>
using namespace std;
template <typename T>
inline void read(T &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
  if (f < 0) x = -x;
}
template <typename T>
inline void write(T x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
int n;
int a[300010];
int num = 0;
int ans = 0;
int stk[300010][35];
long long suml[35][2];
void work(int len, int now) {
  if (len <= 0) return;
  long long gs0 = 0, gs1 = 0, a1 = 0, a2 = 0;
  int top = 0;
  if (now >= 1) top = 0;
  for (int i = 1; i <= len; i++) {
    if ((a[stk[i][now]] >> now) & 1) {
      a1 += gs0;
      ++gs1;
      if (now >= 1) top++, stk[top][now - 1] = stk[i][now];
    } else {
      a2 += gs1;
      ++gs0;
    }
  }
  suml[now][1] += (long long)a1;
  suml[now][0] += (long long)a2;
  if (now == 0) return;
  work(top, now - 1);
  top = 0;
  for (int i = 1; i <= len; i++) {
    if (!((a[stk[i][now]] >> now) & 1)) top++, stk[top][now - 1] = stk[i][now];
  }
  work(top, now - 1);
  return;
}
int main() {
  read(n);
  int x = 0;
  for (int i = 1; i <= n; i++) {
    read(a[i]);
    x = max(x, a[i]);
  }
  int loc = 30;
  while (!((1 << loc) & x) && loc >= 1) loc--;
  int top = 0;
  for (int i = 1; i <= n; i++) {
    top++;
    stk[top][loc] = i;
  }
  work(top, loc);
  long long ans2 = 0ll;
  for (int i = 0; i <= loc; i++) {
    if (suml[i][0] <= suml[i][1]) {
      ans2 += suml[i][0];
    } else {
      ans2 += suml[i][1];
      ans |= (1 << i);
    }
  }
  write(ans2);
  putchar(' ');
  write(ans);
  putchar('\n');
  return 0;
}

```