#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9;
const long long inf64 = 1e18;
const int LOG = 30;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n + 1);
  vector<vector<int>> q[LOG + 1];
  vector<int> perm(n);
  for (int i = 1; i <= n; i++) perm[i - 1] = i;
  for (int i = 1; i <= n; i++) cin >> a[i];
  q[LOG].push_back(perm);
  long long inv = 0, x = 0;
  for (int bit = LOG; bit >= 0; bit--) {
    long long total1 = 0;
    long long total2 = 0;
    for (auto seq : q[bit]) {
      if (seq.empty()) continue;
      vector<int> t0, t1;
      for (int id : seq)
        if ((1 << bit) & a[id])
          t1.push_back(id);
        else
          t0.push_back(id);
      long long hlp1 = 0;
      long long hlp2 = 0;
      for (int j = 0, i = 0; i < (int)t0.size(); i++) {
        while (j < (int)t1.size() && t1[j] < t0[i]) j++;
        hlp1 += j;
      }
      for (int j = 0, i = 0; i < (int)t1.size(); i++) {
        while (j < (int)t0.size() && t0[j] < t1[i]) j++;
        hlp2 += j;
      }
      total1 += hlp1;
      total2 += hlp2;
      if (bit > 0) {
        q[bit - 1].push_back(t0);
        q[bit - 1].push_back(t1);
      }
    }
    if (total2 < total1) {
      inv += total2;
      x |= (1ll << bit);
    } else {
      inv += total1;
    }
  }
  cout << inv << " " << x << "\n";
  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>
using namespace std;
const int inf = 1e9;
const long long inf64 = 1e18;
const int LOG = 30;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n + 1);
  vector<vector<int>> q[LOG + 1];
  vector<int> perm(n);
  for (int i = 1; i <= n; i++) perm[i - 1] = i;
  for (int i = 1; i <= n; i++) cin >> a[i];
  q[LOG].push_back(perm);
  long long inv = 0, x = 0;
  for (int bit = LOG; bit >= 0; bit--) {
    long long total1 = 0;
    long long total2 = 0;
    for (auto seq : q[bit]) {
      if (seq.empty()) continue;
      vector<int> t0, t1;
      for (int id : seq)
        if ((1 << bit) & a[id])
          t1.push_back(id);
        else
          t0.push_back(id);
      long long hlp1 = 0;
      long long hlp2 = 0;
      for (int j = 0, i = 0; i < (int)t0.size(); i++) {
        while (j < (int)t1.size() && t1[j] < t0[i]) j++;
        hlp1 += j;
      }
      for (int j = 0, i = 0; i < (int)t1.size(); i++) {
        while (j < (int)t0.size() && t0[j] < t1[i]) j++;
        hlp2 += j;
      }
      total1 += hlp1;
      total2 += hlp2;
      if (bit > 0) {
        q[bit - 1].push_back(t0);
        q[bit - 1].push_back(t1);
      }
    }
    if (total2 < total1) {
      inv += total2;
      x |= (1ll << bit);
    } else {
      inv += total1;
    }
  }
  cout << inv << " " << x << "\n";
  return 0;
}

```