#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("O3")
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int get_random() {
  static uniform_int_distribution<int> dist(0, 1e9 + 6);
  return dist(rng);
}
template <class T>
void make_unique(T& v) {
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
}
int geti() {
  int x;
  cin >> x;
  return x;
}
long long getll() {
  long long x;
  cin >> x;
  return x;
}
double getd() {
  double x;
  cin >> x;
  return x;
}
const int MAXN = 3e5 + 100;
void solve(int tt) {}
long long a[MAXN], b[MAXN];
int main(int argc, char* argv[]) {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int n;
  cin >> n;
  for (auto i = (0); i < (n); i++) {
    cin >> a[i];
    b[i] = a[i];
  }
  long long res = 0, tot = 0;
  vector<pair<int, int> > itvs({{0, n - 1}});
  long long flag = 1LL << 30;
  long long cflag = flag;
  for (; flag; flag >>= 1) {
    long long res0 = 0;
    long long res1 = 0;
    for (auto itv : itvs) {
      vector<int> tmp;
      int lb = itv.first, ub = itv.second;
      for (auto i = (lb); i < (ub + 1); i++) {
        if (a[i] & flag) tmp.push_back(i);
      }
      for (auto i = (0); i < (tmp.size()); i++) {
        res0 += ub - tmp[i] - (tmp.size() - i - 1);
        res1 += tmp[i] - lb - i;
      }
    }
    if (res1 < res0) res |= flag;
    tot += min(res1, res0);
    cflag |= flag;
    stable_sort(a, a + n, [&](const long long& x, const long long& y) {
      return (x & cflag) < (y & cflag);
    });
    itvs.clear();
    int i = 0, j;
    for (; i < n; i = j) {
      for (j = i; j < n && (cflag & a[j]) == (cflag & a[i]); j++)
        ;
      itvs.push_back({i, j - 1});
    }
  }
  cout << tot << ' ' << res << endl;
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
#pragma GCC optimize("O3")
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int get_random() {
  static uniform_int_distribution<int> dist(0, 1e9 + 6);
  return dist(rng);
}
template <class T>
void make_unique(T& v) {
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
}
int geti() {
  int x;
  cin >> x;
  return x;
}
long long getll() {
  long long x;
  cin >> x;
  return x;
}
double getd() {
  double x;
  cin >> x;
  return x;
}
const int MAXN = 3e5 + 100;
void solve(int tt) {}
long long a[MAXN], b[MAXN];
int main(int argc, char* argv[]) {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int n;
  cin >> n;
  for (auto i = (0); i < (n); i++) {
    cin >> a[i];
    b[i] = a[i];
  }
  long long res = 0, tot = 0;
  vector<pair<int, int> > itvs({{0, n - 1}});
  long long flag = 1LL << 30;
  long long cflag = flag;
  for (; flag; flag >>= 1) {
    long long res0 = 0;
    long long res1 = 0;
    for (auto itv : itvs) {
      vector<int> tmp;
      int lb = itv.first, ub = itv.second;
      for (auto i = (lb); i < (ub + 1); i++) {
        if (a[i] & flag) tmp.push_back(i);
      }
      for (auto i = (0); i < (tmp.size()); i++) {
        res0 += ub - tmp[i] - (tmp.size() - i - 1);
        res1 += tmp[i] - lb - i;
      }
    }
    if (res1 < res0) res |= flag;
    tot += min(res1, res0);
    cflag |= flag;
    stable_sort(a, a + n, [&](const long long& x, const long long& y) {
      return (x & cflag) < (y & cflag);
    });
    itvs.clear();
    int i = 0, j;
    for (; i < n; i = j) {
      for (j = i; j < n && (cflag & a[j]) == (cflag & a[i]); j++)
        ;
      itvs.push_back({i, j - 1});
    }
  }
  cout << tot << ' ' << res << endl;
  return 0;
}

```