#include <bits/stdc++.h>
using namespace std;
const int MAXN = 3e5 + 5;
int n, a[MAXN];
pair<int, int> b[MAXN];
vector<pair<int, int> > inv[100];
int getBit(int x, int i) { return ((x & (1 << i)) > 0); }
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    b[i].first = a[i];
    b[i].second = i;
  }
  a[n] = -1;
  sort(b, b + n);
  int mask = (1 << 31);
  for (int w = 30; w >= 0; --w) {
    int prev = 0;
    for (int i = 1; i <= n; ++i) {
      if (((b[i].first & mask) != (b[i - 1].first & mask))) {
        inv[w].push_back(pair<int, int>(prev, i - 1));
        prev = i;
      }
    }
    if (inv[w].empty()) inv[w].push_back(pair<int, int>(0, n - 1));
    mask += (1 << w);
  }
  int x = 0;
  long long totInversion = 0;
  for (int w = 0; w <= 30; ++w) {
    long long costRemain = 0, costChanged = 0;
    for (pair<int, int> i : inv[w]) {
      int num0 = 0, num1 = 0;
      vector<int> tmp;
      for (int j = i.first; j <= i.second; ++j) {
        tmp.push_back(b[j].second);
      }
      sort(tmp.begin(), tmp.end());
      for (int j : tmp) {
        int idx = j;
        if (getBit(a[idx], w) == 0) {
          costRemain += num1;
          num0++;
        } else {
          costChanged += num0;
          num1++;
        }
      }
    }
    if (costRemain > costChanged) {
      x += (1 << w);
      totInversion += costChanged;
    } else {
      totInversion += costRemain;
    }
  }
  printf("%lld %d\n", totInversion, x);
}

### Prompt

Please provide a cpp coded solution 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 MAXN = 3e5 + 5;
int n, a[MAXN];
pair<int, int> b[MAXN];
vector<pair<int, int> > inv[100];
int getBit(int x, int i) { return ((x & (1 << i)) > 0); }
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    b[i].first = a[i];
    b[i].second = i;
  }
  a[n] = -1;
  sort(b, b + n);
  int mask = (1 << 31);
  for (int w = 30; w >= 0; --w) {
    int prev = 0;
    for (int i = 1; i <= n; ++i) {
      if (((b[i].first & mask) != (b[i - 1].first & mask))) {
        inv[w].push_back(pair<int, int>(prev, i - 1));
        prev = i;
      }
    }
    if (inv[w].empty()) inv[w].push_back(pair<int, int>(0, n - 1));
    mask += (1 << w);
  }
  int x = 0;
  long long totInversion = 0;
  for (int w = 0; w <= 30; ++w) {
    long long costRemain = 0, costChanged = 0;
    for (pair<int, int> i : inv[w]) {
      int num0 = 0, num1 = 0;
      vector<int> tmp;
      for (int j = i.first; j <= i.second; ++j) {
        tmp.push_back(b[j].second);
      }
      sort(tmp.begin(), tmp.end());
      for (int j : tmp) {
        int idx = j;
        if (getBit(a[idx], w) == 0) {
          costRemain += num1;
          num0++;
        } else {
          costChanged += num0;
          num1++;
        }
      }
    }
    if (costRemain > costChanged) {
      x += (1 << w);
      totInversion += costChanged;
    } else {
      totInversion += costRemain;
    }
  }
  printf("%lld %d\n", totInversion, x);
}

```