#include <bits/stdc++.h>
using namespace std;
vector<vector<long long> > vec[2];
vector<long long> x, y;
int main() {
  long long n, bit = 0, ans1 = 0, ans2 = 0, a, b;
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    scanf("%I64d", &a);
    x.push_back(a);
  }
  vec[bit].push_back(x);
  for (long long i = 32; i >= 0; i--) {
    long long bb = (1ll << i), curAns = 0, curAns2 = 0;
    for (auto v : vec[bit]) {
      long long cnt = 0, cnt2 = 0;
      x.clear();
      y.clear();
      for (auto cur : v)
        if (cur & bb) {
          x.push_back(cur);
          cnt++;
          curAns2 += cnt2;
        } else {
          curAns += cnt;
          cnt2++;
          y.push_back(cur);
        }
      if (!x.empty()) vec[bit ^ 1].push_back(x);
      if (!y.empty()) vec[bit ^ 1].push_back(y);
    }
    ans1 += min(curAns, curAns2);
    if (curAns2 < curAns) {
      ans2 += bb;
    }
    vec[bit].clear();
    bit ^= 1;
  }
  cout << ans1 << " " << ans2 << endl;
  return 0;
}

### Prompt

Please formulate a Cpp solution to 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;
vector<vector<long long> > vec[2];
vector<long long> x, y;
int main() {
  long long n, bit = 0, ans1 = 0, ans2 = 0, a, b;
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    scanf("%I64d", &a);
    x.push_back(a);
  }
  vec[bit].push_back(x);
  for (long long i = 32; i >= 0; i--) {
    long long bb = (1ll << i), curAns = 0, curAns2 = 0;
    for (auto v : vec[bit]) {
      long long cnt = 0, cnt2 = 0;
      x.clear();
      y.clear();
      for (auto cur : v)
        if (cur & bb) {
          x.push_back(cur);
          cnt++;
          curAns2 += cnt2;
        } else {
          curAns += cnt;
          cnt2++;
          y.push_back(cur);
        }
      if (!x.empty()) vec[bit ^ 1].push_back(x);
      if (!y.empty()) vec[bit ^ 1].push_back(y);
    }
    ans1 += min(curAns, curAns2);
    if (curAns2 < curAns) {
      ans2 += bb;
    }
    vec[bit].clear();
    bit ^= 1;
  }
  cout << ans1 << " " << ans2 << endl;
  return 0;
}

```