#include <bits/stdc++.h>
using namespace std;
const int N = 30 * 300005;
int cur = 0, g[N][2], sz[N];
long long dp[30][2];
int n;
void ins(int x) {
  int curr = 0;
  for (int i = 29; i >= 0; i--) {
    int b = ((1 << i) & x) ? 1 : 0;
    if (g[curr][!b]) {
      dp[i][b] += sz[g[curr][!b]];
    }
    if (!g[curr][b]) g[curr][b] = ++cur;
    curr = g[curr][b];
    sz[curr]++;
  }
}
int32_t main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    ins(x);
  }
  long long ans = 0, x = 0;
  for (int i = 29; i >= 0; i--) {
    if (dp[i][0] > dp[i][1]) x ^= (1LL << i);
    ans += min(dp[i][0], dp[i][1]);
  }
  printf("%lld %lld\n", ans, x);
}

### 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;
const int N = 30 * 300005;
int cur = 0, g[N][2], sz[N];
long long dp[30][2];
int n;
void ins(int x) {
  int curr = 0;
  for (int i = 29; i >= 0; i--) {
    int b = ((1 << i) & x) ? 1 : 0;
    if (g[curr][!b]) {
      dp[i][b] += sz[g[curr][!b]];
    }
    if (!g[curr][b]) g[curr][b] = ++cur;
    curr = g[curr][b];
    sz[curr]++;
  }
}
int32_t main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    ins(x);
  }
  long long ans = 0, x = 0;
  for (int i = 29; i >= 0; i--) {
    if (dp[i][0] > dp[i][1]) x ^= (1LL << i);
    ans += min(dp[i][0], dp[i][1]);
  }
  printf("%lld %lld\n", ans, x);
}

```