#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5;
const int p = 1e9 + 7;
long long qpow(long long a, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) res = res * a % p;
    a = a * a % p;
    n >>= 1;
  }
  return res;
}
int a[N];
int b[N];
long long sum[2][31];
void dfs(int l, int r, int x) {
  if (l > r || x < 0) return;
  int cnt1 = 0;
  int cnt2 = 0;
  for (int i = l; i <= r; i++) {
    if (a[i] & (1 << x)) {
      cnt1++;
      sum[0][x] += cnt2;
    } else {
      cnt2++;
      sum[1][x] += cnt1;
    }
  }
  int now = l - 1;
  for (int i = l; i <= r; i++) {
    if (a[i] & (1 << x)) {
      b[++now] = a[i];
    }
  }
  int m = now;
  for (int i = l; i <= r; i++) {
    if (!(a[i] & (1 << x))) {
      b[++now] = a[i];
    }
  }
  for (int i = l; i <= r; i++) {
    a[i] = b[i];
  }
  dfs(l, m, x - 1);
  dfs(m + 1, r, x - 1);
}
void work() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  dfs(1, n, 30);
  long long ans = 0;
  int x = 0;
  for (int i = 0; i <= 30; i++) {
    if (sum[0][i] < sum[1][i]) {
      x |= (1 << i);
    }
    ans += min(sum[0][i], sum[1][i]);
  }
  printf("%lld %d\n", ans, x);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  while (T--) {
    work();
  }
}

### Prompt

Develop a solution in CPP 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;
const int N = 3e5 + 5;
const int p = 1e9 + 7;
long long qpow(long long a, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) res = res * a % p;
    a = a * a % p;
    n >>= 1;
  }
  return res;
}
int a[N];
int b[N];
long long sum[2][31];
void dfs(int l, int r, int x) {
  if (l > r || x < 0) return;
  int cnt1 = 0;
  int cnt2 = 0;
  for (int i = l; i <= r; i++) {
    if (a[i] & (1 << x)) {
      cnt1++;
      sum[0][x] += cnt2;
    } else {
      cnt2++;
      sum[1][x] += cnt1;
    }
  }
  int now = l - 1;
  for (int i = l; i <= r; i++) {
    if (a[i] & (1 << x)) {
      b[++now] = a[i];
    }
  }
  int m = now;
  for (int i = l; i <= r; i++) {
    if (!(a[i] & (1 << x))) {
      b[++now] = a[i];
    }
  }
  for (int i = l; i <= r; i++) {
    a[i] = b[i];
  }
  dfs(l, m, x - 1);
  dfs(m + 1, r, x - 1);
}
void work() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  dfs(1, n, 30);
  long long ans = 0;
  int x = 0;
  for (int i = 0; i <= 30; i++) {
    if (sum[0][i] < sum[1][i]) {
      x |= (1 << i);
    }
    ans += min(sum[0][i], sum[1][i]);
  }
  printf("%lld %d\n", ans, x);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  while (T--) {
    work();
  }
}

```