#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = x * 10 + c - '0', c = getchar();
  }
  return x * f;
}
int T, n, cnt, a[6000010], sz[6000010];
long long f[2][110];
struct TRIE {
  int ch[2];
} t[6000010];
void add(int x) {
  int now = 0;
  for (int i = 30; i >= 0; i--) {
    int k = (x >> i) & 1;
    if (!t[now].ch[k]) t[now].ch[k] = ++cnt;
    f[k ^ 1][i] += sz[t[now].ch[k ^ 1]];
    now = t[now].ch[k], sz[now]++;
  }
}
void solve() {
  n = read();
  long long ans = 0, qwq = 0;
  for (int i = 1; i <= n; i++) a[i] = read(), add(a[i]);
  for (int i = 30; i >= 0; i--) {
    if (f[0][i] < f[1][i]) {
      qwq += f[0][i];
      ans += 1 << (i);
    } else
      qwq += f[1][i];
  }
  printf("%lld %lld", qwq, ans);
}
int main() {
  solve();
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = x * 10 + c - '0', c = getchar();
  }
  return x * f;
}
int T, n, cnt, a[6000010], sz[6000010];
long long f[2][110];
struct TRIE {
  int ch[2];
} t[6000010];
void add(int x) {
  int now = 0;
  for (int i = 30; i >= 0; i--) {
    int k = (x >> i) & 1;
    if (!t[now].ch[k]) t[now].ch[k] = ++cnt;
    f[k ^ 1][i] += sz[t[now].ch[k ^ 1]];
    now = t[now].ch[k], sz[now]++;
  }
}
void solve() {
  n = read();
  long long ans = 0, qwq = 0;
  for (int i = 1; i <= n; i++) a[i] = read(), add(a[i]);
  for (int i = 30; i >= 0; i--) {
    if (f[0][i] < f[1][i]) {
      qwq += f[0][i];
      ans += 1 << (i);
    } else
      qwq += f[1][i];
  }
  printf("%lld %lld", qwq, ans);
}
int main() {
  solve();
  return 0;
}

```