#include <bits/stdc++.h>
using namespace std;
const int MAXN = 3e5;
int N, A[MAXN + 10];
int X;
long long ans;
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%d", &A[i]);
  vector<vector<int>> V;
  V.push_back(vector<int>(A + 1, A + N + 1));
  for (int i = 29; i >= 0; i--) {
    long long v0 = 0, v1 = 0;
    for (auto it : V) {
      int cnt = 0;
      for (auto jt : it) {
        if (jt & (1 << i))
          cnt++;
        else
          v0 += cnt;
      }
      cnt = 0;
      for (auto jt : it) {
        if (jt & (1 << i))
          v1 += cnt;
        else
          cnt++;
      }
    }
    vector<vector<int>> V2;
    for (auto it : V) {
      vector<int> VV1, VV2;
      for (auto jt : it) {
        if (jt & (1 << i))
          VV1.push_back(jt);
        else
          VV2.push_back(jt);
      }
      if (!VV1.empty()) V2.push_back(VV1);
      if (!VV2.empty()) V2.push_back(VV2);
    }
    V = V2;
    if (v0 <= v1) {
      ans += v0;
    } else {
      ans += v1;
      X |= (1 << i);
    }
  }
  printf("%lld %d\n", ans, X);
}

### Prompt

Develop a solution in CPP to the problem described below:

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 MAXN = 3e5;
int N, A[MAXN + 10];
int X;
long long ans;
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%d", &A[i]);
  vector<vector<int>> V;
  V.push_back(vector<int>(A + 1, A + N + 1));
  for (int i = 29; i >= 0; i--) {
    long long v0 = 0, v1 = 0;
    for (auto it : V) {
      int cnt = 0;
      for (auto jt : it) {
        if (jt & (1 << i))
          cnt++;
        else
          v0 += cnt;
      }
      cnt = 0;
      for (auto jt : it) {
        if (jt & (1 << i))
          v1 += cnt;
        else
          cnt++;
      }
    }
    vector<vector<int>> V2;
    for (auto it : V) {
      vector<int> VV1, VV2;
      for (auto jt : it) {
        if (jt & (1 << i))
          VV1.push_back(jt);
        else
          VV2.push_back(jt);
      }
      if (!VV1.empty()) V2.push_back(VV1);
      if (!VV2.empty()) V2.push_back(VV2);
    }
    V = V2;
    if (v0 <= v1) {
      ans += v0;
    } else {
      ans += v1;
      X |= (1 << i);
    }
  }
  printf("%lld %d\n", ans, X);
}

```