#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 10;
long long ans, res;
int L[N], R[N];
int n, a[N];
long long dp[40][2];
void solve(int d, int l, int r) {
  if (l >= r) return;
  if (d == -1) return;
  int nl = 0, nr = 0;
  long long tmp1 = 0, tmp2 = 0;
  ;
  for (int i = l; i <= r; i++) {
    if (a[i] >> d & 1)
      L[++nl] = a[i], tmp1 += nr;
    else
      R[++nr] = a[i], tmp2 += nl;
  }
  dp[d][1] += tmp1;
  dp[d][0] += tmp2;
  int x = 1, y = 1;
  for (int i = l; i <= r; i++) {
    if (x <= nl)
      a[i] = L[x++];
    else
      a[i] = R[y++];
  }
  solve(d - 1, l, l + nl - 1);
  solve(d - 1, l + nl, r);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  solve(30, 1, n);
  for (int i = 0; i <= 30; i++) {
    if (dp[i][1] < dp[i][0])
      ans += dp[i][1], res |= (1 << i);
    else
      ans += dp[i][0];
  }
  printf("%lld %lld\n", ans, res);
}

### 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 + 10;
long long ans, res;
int L[N], R[N];
int n, a[N];
long long dp[40][2];
void solve(int d, int l, int r) {
  if (l >= r) return;
  if (d == -1) return;
  int nl = 0, nr = 0;
  long long tmp1 = 0, tmp2 = 0;
  ;
  for (int i = l; i <= r; i++) {
    if (a[i] >> d & 1)
      L[++nl] = a[i], tmp1 += nr;
    else
      R[++nr] = a[i], tmp2 += nl;
  }
  dp[d][1] += tmp1;
  dp[d][0] += tmp2;
  int x = 1, y = 1;
  for (int i = l; i <= r; i++) {
    if (x <= nl)
      a[i] = L[x++];
    else
      a[i] = R[y++];
  }
  solve(d - 1, l, l + nl - 1);
  solve(d - 1, l + nl, r);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  solve(30, 1, n);
  for (int i = 0; i <= 30; i++) {
    if (dp[i][1] < dp[i][0])
      ans += dp[i][1], res |= (1 << i);
    else
      ans += dp[i][0];
  }
  printf("%lld %lld\n", ans, res);
}

```