#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5;
long long calc(vector<int> a, int &A1, int &A2) {}
int n, a[N], ans;
long long Ans;
vector<vector<int> > f, g;
void solve(int z) {
  long long ans0 = 0, ans1 = 0;
  int l = f.size();
  for (int k = 0; k <= l - 1; k++) {
    int len = f[k].size(), sum0 = 0, sum1 = 0;
    vector<int> b(len);
    for (int i = 0; i <= len - 1; i++) b[i] = ((f[k][i] >> z) & 1);
    for (int i = 0; i <= len - 1; i++)
      if (b[i])
        sum1++;
      else
        sum0++;
    for (int i = 0; i <= len - 1; i++)
      if (b[i] == 0)
        sum0--;
      else
        ans0 += sum0;
    sum0 = len - sum1;
    swap(sum0, sum1);
    for (int i = 0; i <= len - 1; i++)
      if (b[i] == 1)
        sum0--;
      else
        ans1 += sum0;
  }
  g.clear();
  if (ans1 < ans0) {
    ans ^= 1 << z;
    Ans += ans1;
  } else {
    Ans += ans0;
  }
  for (int i = 0; i <= l - 1; i++) {
    vector<int> d = f[i], d0(0), d1(0);
    for (int j = 0; j <= (int)d.size() - 1; j++) {
      if (((d[j] ^ ans) >> z) & 1)
        d1.push_back(d[j]);
      else
        d0.push_back(d[j]);
    }
    if (d0.size()) g.push_back(d0);
    if (d1.size()) g.push_back(d1);
  }
  f = g;
  g.clear();
  if (z > 0) solve(z - 1);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  vector<int> h;
  for (int i = 1; i <= n; i++) h.push_back(a[i]);
  f.push_back(h);
  solve(30);
  printf("%lld %d\n", Ans, ans);
  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 N = 3e5 + 5;
long long calc(vector<int> a, int &A1, int &A2) {}
int n, a[N], ans;
long long Ans;
vector<vector<int> > f, g;
void solve(int z) {
  long long ans0 = 0, ans1 = 0;
  int l = f.size();
  for (int k = 0; k <= l - 1; k++) {
    int len = f[k].size(), sum0 = 0, sum1 = 0;
    vector<int> b(len);
    for (int i = 0; i <= len - 1; i++) b[i] = ((f[k][i] >> z) & 1);
    for (int i = 0; i <= len - 1; i++)
      if (b[i])
        sum1++;
      else
        sum0++;
    for (int i = 0; i <= len - 1; i++)
      if (b[i] == 0)
        sum0--;
      else
        ans0 += sum0;
    sum0 = len - sum1;
    swap(sum0, sum1);
    for (int i = 0; i <= len - 1; i++)
      if (b[i] == 1)
        sum0--;
      else
        ans1 += sum0;
  }
  g.clear();
  if (ans1 < ans0) {
    ans ^= 1 << z;
    Ans += ans1;
  } else {
    Ans += ans0;
  }
  for (int i = 0; i <= l - 1; i++) {
    vector<int> d = f[i], d0(0), d1(0);
    for (int j = 0; j <= (int)d.size() - 1; j++) {
      if (((d[j] ^ ans) >> z) & 1)
        d1.push_back(d[j]);
      else
        d0.push_back(d[j]);
    }
    if (d0.size()) g.push_back(d0);
    if (d1.size()) g.push_back(d1);
  }
  f = g;
  g.clear();
  if (z > 0) solve(z - 1);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  vector<int> h;
  for (int i = 1; i <= n; i++) h.push_back(a[i]);
  f.push_back(h);
  solve(30);
  printf("%lld %d\n", Ans, ans);
  return 0;
}

```