#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long sum = 0, ff = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') ff = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') sum = sum * 10 + ch - '0', ch = getchar();
  return sum * ff;
}
const int N = 5e6 + 7;
const int maxbit = 31;
int n, cnt, x;
long long f[N][3], num;
struct tree {
  int son[3];
} t[N];
vector<int> v[N];
void insert(int val, int pos) {
  int now = 0;
  for (int i = maxbit; i >= 0; i--) {
    bool bit = ((val >> i) & 1);
    if (!t[now].son[bit]) t[now].son[bit] = ++cnt;
    now = t[now].son[bit], v[now].push_back(pos);
  }
}
void dfs(int now, int bit) {
  int l = t[now].son[0], r = t[now].son[1];
  if (l) dfs(l, bit - 1);
  if (r) dfs(r, bit - 1);
  if (!l && !r) return;
  long long sum = 0;
  int pos = 0;
  for (int i = 0; i < v[l].size(); i++) {
    int x = v[l][i];
    while (pos < v[r].size() && v[r][pos] < x) pos++;
    sum += pos;
  }
  f[bit][0] += sum, f[bit][1] += 1ll * v[l].size() * v[r].size() - sum;
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    int val = read();
    insert(val, i);
  }
  dfs(0, maxbit);
  for (int i = 0; i <= maxbit; i++) {
    num += min(f[i][0], f[i][1]);
    if (f[i][1] < f[i][0]) x += (1 << i);
  }
  cout << num << " " << x;
  return 0;
}

### Prompt

Your task is to create 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;
inline long long read() {
  long long sum = 0, ff = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') ff = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') sum = sum * 10 + ch - '0', ch = getchar();
  return sum * ff;
}
const int N = 5e6 + 7;
const int maxbit = 31;
int n, cnt, x;
long long f[N][3], num;
struct tree {
  int son[3];
} t[N];
vector<int> v[N];
void insert(int val, int pos) {
  int now = 0;
  for (int i = maxbit; i >= 0; i--) {
    bool bit = ((val >> i) & 1);
    if (!t[now].son[bit]) t[now].son[bit] = ++cnt;
    now = t[now].son[bit], v[now].push_back(pos);
  }
}
void dfs(int now, int bit) {
  int l = t[now].son[0], r = t[now].son[1];
  if (l) dfs(l, bit - 1);
  if (r) dfs(r, bit - 1);
  if (!l && !r) return;
  long long sum = 0;
  int pos = 0;
  for (int i = 0; i < v[l].size(); i++) {
    int x = v[l][i];
    while (pos < v[r].size() && v[r][pos] < x) pos++;
    sum += pos;
  }
  f[bit][0] += sum, f[bit][1] += 1ll * v[l].size() * v[r].size() - sum;
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    int val = read();
    insert(val, i);
  }
  dfs(0, maxbit);
  for (int i = 0; i <= maxbit; i++) {
    num += min(f[i][0], f[i][1]);
    if (f[i][1] < f[i][0]) x += (1 << i);
  }
  cout << num << " " << x;
  return 0;
}

```