#include <bits/stdc++.h>
using namespace std;
const int DIV = 1000000007;
class BIT {
 private:
  vector<int> t;

 public:
  BIT(int n) : t(n, 0) {}
  int sum(int i) {
    int result = 0;
    while (i >= 0) {
      result += t[i];
      i = (i & (i + 1)) - 1;
    }
    return result;
  }
  int sum(int l, int r) {
    int result = sum(r);
    if (l > 0) result -= sum(l - 1);
    return result;
  }
  void add(int i, int value) {
    while (i < t.size()) {
      t[i] += value;
      i = i | (i + 1);
    }
  }
};
int64_t getInvCount(int n, vector<int> &a, vector<pair<int, int>> &b,
                    vector<int> &c, int x) {
  for (int i = 0; i < n; i++) {
    b[i].first = a[i] ^ x;
    b[i].second = i;
  }
  sort(b.begin(), b.end());
  int MAX = 0;
  for (int i = 0; i < n; i++) {
    if (i > 0 && b[i].first > b[i - 1].first) MAX++;
    c[b[i].second] = MAX;
  }
  BIT bit(MAX + 1);
  int64_t result = 0;
  for (int i = n - 1; i >= 0; i--) {
    if (c[i] > 0) result += bit.sum(c[i] - 1);
    bit.add(c[i], 1);
  }
  return result;
}
void solve(istream &in, int test) {
  int n;
  in >> n;
  vector<int> a(n);
  int MAX = 30;
  for (int i = 0; i < n; i++) {
    in >> a[i];
  }
  vector<pair<int, int>> b(n);
  vector<int> c(n);
  int64_t res0 = getInvCount(n, a, b, c, 0);
  int x = 0;
  for (int i = 0; i < MAX; i++) {
    int curX = 1 << i;
    int64_t curRes = getInvCount(n, a, b, c, curX);
    if (curRes < res0) x ^= curX;
  }
  int64_t res = getInvCount(n, a, b, c, x);
  printf("%I64d %d\n", res, x);
}
int main(int argc, char *argv[]) {
  istream &f = cin;
  int t = 1;
  for (int test = 0; test < t; test++) {
    solve(f, test);
  }
  return 0;
}

### 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 DIV = 1000000007;
class BIT {
 private:
  vector<int> t;

 public:
  BIT(int n) : t(n, 0) {}
  int sum(int i) {
    int result = 0;
    while (i >= 0) {
      result += t[i];
      i = (i & (i + 1)) - 1;
    }
    return result;
  }
  int sum(int l, int r) {
    int result = sum(r);
    if (l > 0) result -= sum(l - 1);
    return result;
  }
  void add(int i, int value) {
    while (i < t.size()) {
      t[i] += value;
      i = i | (i + 1);
    }
  }
};
int64_t getInvCount(int n, vector<int> &a, vector<pair<int, int>> &b,
                    vector<int> &c, int x) {
  for (int i = 0; i < n; i++) {
    b[i].first = a[i] ^ x;
    b[i].second = i;
  }
  sort(b.begin(), b.end());
  int MAX = 0;
  for (int i = 0; i < n; i++) {
    if (i > 0 && b[i].first > b[i - 1].first) MAX++;
    c[b[i].second] = MAX;
  }
  BIT bit(MAX + 1);
  int64_t result = 0;
  for (int i = n - 1; i >= 0; i--) {
    if (c[i] > 0) result += bit.sum(c[i] - 1);
    bit.add(c[i], 1);
  }
  return result;
}
void solve(istream &in, int test) {
  int n;
  in >> n;
  vector<int> a(n);
  int MAX = 30;
  for (int i = 0; i < n; i++) {
    in >> a[i];
  }
  vector<pair<int, int>> b(n);
  vector<int> c(n);
  int64_t res0 = getInvCount(n, a, b, c, 0);
  int x = 0;
  for (int i = 0; i < MAX; i++) {
    int curX = 1 << i;
    int64_t curRes = getInvCount(n, a, b, c, curX);
    if (curRes < res0) x ^= curX;
  }
  int64_t res = getInvCount(n, a, b, c, x);
  printf("%I64d %d\n", res, x);
}
int main(int argc, char *argv[]) {
  istream &f = cin;
  int t = 1;
  for (int test = 0; test < t; test++) {
    solve(f, test);
  }
  return 0;
}

```