#include <bits/stdc++.h>
using namespace std;
const long long N = 3e5 + 50;
const long long mod = 998244353;
long long level[31][2];
void func(vector<long long> a, long long i) {
  vector<long long> zeros, ones;
  long long cnt0 = 0, cnt1 = 0;
  for (auto x : a) {
    if (x & (1ll << i)) {
      level[i][1] += cnt0;
      cnt1++;
      ones.push_back(x);
    } else {
      level[i][0] += cnt1;
      cnt0++;
      zeros.push_back(x);
    }
  }
  if (i > 0) {
    if (ones.empty() == false) func(ones, i - 1);
    if (zeros.empty() == false) func(zeros, i - 1);
  }
}
void fun(const vector<long long>& a, long long i) {
  vector<long long> zeros, ones;
  long long cnt0 = 0, cnt1 = 0;
  for (auto x : a) {
    if (x & (1ll << i)) {
      level[i][1] += cnt0;
      cnt1++;
      ones.push_back(x);
    } else {
      level[i][0] += cnt1;
      cnt0++;
      zeros.push_back(x);
    }
  }
  if (i > 0) {
    if (ones.empty() == false) fun(ones, i - 1);
    if (zeros.empty() == false) fun(zeros, i - 1);
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  srand(time(NULL));
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  clock_t t = clock();
  long long x = 0;
  t = clock();
  fun(a, 30);
  x = 0;
  long long summ = 0;
  for (long long i = 0; i < 31; i++) {
    if (level[i][1] < level[i][0])
      x |= (1 << i), summ += level[i][1];
    else
      summ += level[i][0];
    level[i][1] = level[i][0] = 0;
  }
  cout << summ << ' ' << x << ' ';
}

### 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;
const long long N = 3e5 + 50;
const long long mod = 998244353;
long long level[31][2];
void func(vector<long long> a, long long i) {
  vector<long long> zeros, ones;
  long long cnt0 = 0, cnt1 = 0;
  for (auto x : a) {
    if (x & (1ll << i)) {
      level[i][1] += cnt0;
      cnt1++;
      ones.push_back(x);
    } else {
      level[i][0] += cnt1;
      cnt0++;
      zeros.push_back(x);
    }
  }
  if (i > 0) {
    if (ones.empty() == false) func(ones, i - 1);
    if (zeros.empty() == false) func(zeros, i - 1);
  }
}
void fun(const vector<long long>& a, long long i) {
  vector<long long> zeros, ones;
  long long cnt0 = 0, cnt1 = 0;
  for (auto x : a) {
    if (x & (1ll << i)) {
      level[i][1] += cnt0;
      cnt1++;
      ones.push_back(x);
    } else {
      level[i][0] += cnt1;
      cnt0++;
      zeros.push_back(x);
    }
  }
  if (i > 0) {
    if (ones.empty() == false) fun(ones, i - 1);
    if (zeros.empty() == false) fun(zeros, i - 1);
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  srand(time(NULL));
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  clock_t t = clock();
  long long x = 0;
  t = clock();
  fun(a, 30);
  x = 0;
  long long summ = 0;
  for (long long i = 0; i < 31; i++) {
    if (level[i][1] < level[i][0])
      x |= (1 << i), summ += level[i][1];
    else
      summ += level[i][0];
    level[i][1] = level[i][0] = 0;
  }
  cout << summ << ' ' << x << ' ';
}

```