#include <bits/stdc++.h>
using namespace std;
const long long N = 3e6 + 1, M = 2e2 + 1, INF = 2e9, MOD = 1e9 + 7, L = 20,
                NN = 1e9 + 1;
long long mas[N];
long long col[N];
long long k0[N];
long long k1[N];
bool used[N];
long long nextcol[N];
long long fre = 2;
void solve() {
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> mas[i];
    col[i] = 1;
  }
  long long inv = 0;
  long long ans = 0;
  for (long long bit = 40; bit >= 0; bit--) {
    long long inv0 = 0, inv1 = 0;
    for (long long i = 0; i < n; i++) {
      if ((mas[i] & (1LL << bit)) == 0) {
        inv0 += k1[col[i]];
        k0[col[i]]++;
      } else {
        inv1 += k0[col[i]];
        k1[col[i]]++;
        if (used[col[i]]) {
          col[i] = nextcol[col[i]];
        } else {
          used[col[i]] = 1;
          nextcol[col[i]] = fre;
          col[i] = fre;
          fre++;
        }
      }
    }
    for (long long i = 0; i < fre; i++) {
      used[i] = 0;
      k0[i] = 0;
      k1[i] = 0;
    }
    if (inv0 <= inv1) {
      inv += inv0;
    } else {
      inv += inv1;
      ans += (1 << bit);
    }
  }
  cout << inv << ' ' << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) solve();
}

### Prompt

Create a solution in cpp for 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 long long N = 3e6 + 1, M = 2e2 + 1, INF = 2e9, MOD = 1e9 + 7, L = 20,
                NN = 1e9 + 1;
long long mas[N];
long long col[N];
long long k0[N];
long long k1[N];
bool used[N];
long long nextcol[N];
long long fre = 2;
void solve() {
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> mas[i];
    col[i] = 1;
  }
  long long inv = 0;
  long long ans = 0;
  for (long long bit = 40; bit >= 0; bit--) {
    long long inv0 = 0, inv1 = 0;
    for (long long i = 0; i < n; i++) {
      if ((mas[i] & (1LL << bit)) == 0) {
        inv0 += k1[col[i]];
        k0[col[i]]++;
      } else {
        inv1 += k0[col[i]];
        k1[col[i]]++;
        if (used[col[i]]) {
          col[i] = nextcol[col[i]];
        } else {
          used[col[i]] = 1;
          nextcol[col[i]] = fre;
          col[i] = fre;
          fre++;
        }
      }
    }
    for (long long i = 0; i < fre; i++) {
      used[i] = 0;
      k0[i] = 0;
      k1[i] = 0;
    }
    if (inv0 <= inv1) {
      inv += inv0;
    } else {
      inv += inv1;
      ans += (1 << bit);
    }
  }
  cout << inv << ' ' << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) solve();
}

```