#include <bits/stdc++.h>
using namespace std;
const int maxn = 4e5 + 5;
int n, m, k;
int a[maxn], c[maxn], b[maxn];
long long ans, tmp, cnt, sum;
int flag;
char s[maxn];
bool ok[maxn];
template <typename T>
inline void read(T &X) {
  X = 0;
  int w = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    w |= ch == '-';
    ch = getchar();
  }
  while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
  if (w) X = -X;
}
void Merge(int begin, int mid, int end) {
  int i = begin, j = mid + 1, pos = begin;
  while (i <= mid && j <= end) {
    if (a[i] <= a[j]) {
      b[pos++] = a[i++];
    } else {
      b[pos++] = a[j++];
      sum += mid - i + 1;
    }
  }
  while (i <= mid) b[pos++] = a[i++];
  while (j <= end) b[pos++] = a[j++];
  for (int i = begin, j = begin; i <= end; i++, j++) a[i] = b[j];
}
void Sort(int begin, int end) {
  if (begin < end) {
    int mid = (begin + end) / 2;
    Sort(begin, mid);
    Sort(mid + 1, end);
    Merge(begin, mid, end);
  }
}
void solve() {
  read(n);
  for (int i = (1); i <= (n); i++) {
    read(c[i]);
  }
  int x = 0;
  int tmp = 0;
  for (int i = (30); i >= (0); i--) {
    tmp |= (1 << i);
    for (int j = (1); j <= (n); j++) a[j] = (c[j] & tmp);
    sum = 0;
    Sort(1, n);
    long long sum1 = sum;
    sum = 0;
    for (int j = (1); j <= (n); j++) a[j] = (c[j] & tmp) ^ (1 << i);
    Sort(1, n);
    if (sum1 <= sum) continue;
    x |= (1 << i);
    for (int j = (1); j <= (n); j++) c[j] ^= (1 << i);
  }
  sum = 0;
  for (int i = (1); i <= (n); i++) a[i] = c[i];
  Sort(1, n);
  printf("%lld %d\n", sum, x);
}
int main() {
  int T = 1, cas = 1;
  while (T--) {
    solve();
  }
  return 0;
}

### Prompt

Your challenge is to write 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>
using namespace std;
const int maxn = 4e5 + 5;
int n, m, k;
int a[maxn], c[maxn], b[maxn];
long long ans, tmp, cnt, sum;
int flag;
char s[maxn];
bool ok[maxn];
template <typename T>
inline void read(T &X) {
  X = 0;
  int w = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    w |= ch == '-';
    ch = getchar();
  }
  while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
  if (w) X = -X;
}
void Merge(int begin, int mid, int end) {
  int i = begin, j = mid + 1, pos = begin;
  while (i <= mid && j <= end) {
    if (a[i] <= a[j]) {
      b[pos++] = a[i++];
    } else {
      b[pos++] = a[j++];
      sum += mid - i + 1;
    }
  }
  while (i <= mid) b[pos++] = a[i++];
  while (j <= end) b[pos++] = a[j++];
  for (int i = begin, j = begin; i <= end; i++, j++) a[i] = b[j];
}
void Sort(int begin, int end) {
  if (begin < end) {
    int mid = (begin + end) / 2;
    Sort(begin, mid);
    Sort(mid + 1, end);
    Merge(begin, mid, end);
  }
}
void solve() {
  read(n);
  for (int i = (1); i <= (n); i++) {
    read(c[i]);
  }
  int x = 0;
  int tmp = 0;
  for (int i = (30); i >= (0); i--) {
    tmp |= (1 << i);
    for (int j = (1); j <= (n); j++) a[j] = (c[j] & tmp);
    sum = 0;
    Sort(1, n);
    long long sum1 = sum;
    sum = 0;
    for (int j = (1); j <= (n); j++) a[j] = (c[j] & tmp) ^ (1 << i);
    Sort(1, n);
    if (sum1 <= sum) continue;
    x |= (1 << i);
    for (int j = (1); j <= (n); j++) c[j] ^= (1 << i);
  }
  sum = 0;
  for (int i = (1); i <= (n); i++) a[i] = c[i];
  Sort(1, n);
  printf("%lld %d\n", sum, x);
}
int main() {
  int T = 1, cas = 1;
  while (T--) {
    solve();
  }
  return 0;
}

```