#include <bits/stdc++.h>
const int maxn = 3e5 + 7, maxm = 1e7 + 10;
using namespace std;
int maxs = 0, k, tot = 0;
int t[35][maxn], sum[maxm][2];
long long sumsum = 0, ans = 0;
int b[maxm], a[maxn];
int main() {
  int tot1 = 0;
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    maxs = max(maxs, a[i]);
  }
  k = 33;
  for (int i = k; i >= 0; i--) {
    long long sum0 = 0, sum1 = 0;
    for (int j = 0; j <= tot; j++) sum[j][0] = sum[j][1] = b[j] = 0;
    long long v = 1LL << i;
    for (int j = n; j >= 1; j--) {
      long long num = (v & a[j]);
      if (num) {
        if (!b[t[i + 1][j]]) {
          b[t[i + 1][j]] = ++tot;
          t[i][j] = tot;
        } else
          t[i][j] = b[t[i + 1][j]];
        sum0 += sum[t[i + 1][j]][0];
        sum[t[i + 1][j]][1]++;
      } else {
        t[i][j] = t[i + 1][j];
        sum1 += sum[t[i + 1][j]][1];
        sum[t[i + 1][j]][0]++;
      }
    }
    if (sum0 <= sum1) {
      sumsum += sum0;
    } else {
      sumsum += sum1;
      ans += v;
    }
  }
  cout << sumsum << " " << ans << endl;
  return 0;
}

### Prompt

In CPP, your task is to solve 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>
const int maxn = 3e5 + 7, maxm = 1e7 + 10;
using namespace std;
int maxs = 0, k, tot = 0;
int t[35][maxn], sum[maxm][2];
long long sumsum = 0, ans = 0;
int b[maxm], a[maxn];
int main() {
  int tot1 = 0;
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    maxs = max(maxs, a[i]);
  }
  k = 33;
  for (int i = k; i >= 0; i--) {
    long long sum0 = 0, sum1 = 0;
    for (int j = 0; j <= tot; j++) sum[j][0] = sum[j][1] = b[j] = 0;
    long long v = 1LL << i;
    for (int j = n; j >= 1; j--) {
      long long num = (v & a[j]);
      if (num) {
        if (!b[t[i + 1][j]]) {
          b[t[i + 1][j]] = ++tot;
          t[i][j] = tot;
        } else
          t[i][j] = b[t[i + 1][j]];
        sum0 += sum[t[i + 1][j]][0];
        sum[t[i + 1][j]][1]++;
      } else {
        t[i][j] = t[i + 1][j];
        sum1 += sum[t[i + 1][j]][1];
        sum[t[i + 1][j]][0]++;
      }
    }
    if (sum0 <= sum1) {
      sumsum += sum0;
    } else {
      sumsum += sum1;
      ans += v;
    }
  }
  cout << sumsum << " " << ans << endl;
  return 0;
}

```