#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
template <typename T1, typename T2>
inline void chkmin(T1& x, const T2& y) {
  if (y < x) x = y;
}
template <typename T1, typename T2>
inline void chkmax(T1& x, const T2& y) {
  if (x < y) x = y;
}
const int N = 3e5 + 10;
int n;
int a[N];
void read() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
}
int fenwick[N];
vector<int> coord;
void upd(int pos) {
  for (; pos < coord.size(); pos |= pos + 1) ++fenwick[pos];
}
ll get(int r) {
  ll ans = 0;
  for (; r >= 0; r = (r & (r + 1)) - 1) {
    ans += fenwick[r];
  }
  return ans;
}
ll calcInv(int x) {
  fill(fenwick, fenwick + coord.size(), 0);
  sort((coord).begin(), (coord).end(),
       [&](int a, int b) { return (a ^ x) < (b ^ x); });
  ll ans = 0;
  for (int i = 0; i < n; ++i) {
    int pos =
        lower_bound((coord).begin(), (coord).end(), a[i],
                    [&](int myA, int myB) { return (myA ^ x) < (myB ^ x); }) -
        coord.begin();
    ans += i - get(pos);
    upd(pos);
  }
  return ans;
}
ll ansInv;
int ansX;
void run() {
  for (int i = 0; i < n; ++i) {
    coord.push_back(a[i]);
  }
  sort((coord).begin(), (coord).end());
  coord.resize(unique((coord).begin(), (coord).end()) - coord.begin());
  ansInv = calcInv(0);
  ansX = 0;
  for (int bit = 29; bit >= 0; --bit) {
    ll put = calcInv(ansX ^ (1 << bit));
    ll notPut = ansInv;
    if (put < notPut) {
      ansX ^= (1 << bit);
    }
    ansInv = min(put, notPut);
  }
}
void write() { cout << ansInv << " " << ansX << "\n"; }
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0), cout.precision(20),
      cout.setf(ios::fixed);
  read();
  run();
  write();
  return 0;
}

### Prompt

Your challenge is to write 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;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
template <typename T1, typename T2>
inline void chkmin(T1& x, const T2& y) {
  if (y < x) x = y;
}
template <typename T1, typename T2>
inline void chkmax(T1& x, const T2& y) {
  if (x < y) x = y;
}
const int N = 3e5 + 10;
int n;
int a[N];
void read() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
}
int fenwick[N];
vector<int> coord;
void upd(int pos) {
  for (; pos < coord.size(); pos |= pos + 1) ++fenwick[pos];
}
ll get(int r) {
  ll ans = 0;
  for (; r >= 0; r = (r & (r + 1)) - 1) {
    ans += fenwick[r];
  }
  return ans;
}
ll calcInv(int x) {
  fill(fenwick, fenwick + coord.size(), 0);
  sort((coord).begin(), (coord).end(),
       [&](int a, int b) { return (a ^ x) < (b ^ x); });
  ll ans = 0;
  for (int i = 0; i < n; ++i) {
    int pos =
        lower_bound((coord).begin(), (coord).end(), a[i],
                    [&](int myA, int myB) { return (myA ^ x) < (myB ^ x); }) -
        coord.begin();
    ans += i - get(pos);
    upd(pos);
  }
  return ans;
}
ll ansInv;
int ansX;
void run() {
  for (int i = 0; i < n; ++i) {
    coord.push_back(a[i]);
  }
  sort((coord).begin(), (coord).end());
  coord.resize(unique((coord).begin(), (coord).end()) - coord.begin());
  ansInv = calcInv(0);
  ansX = 0;
  for (int bit = 29; bit >= 0; --bit) {
    ll put = calcInv(ansX ^ (1 << bit));
    ll notPut = ansInv;
    if (put < notPut) {
      ansX ^= (1 << bit);
    }
    ansInv = min(put, notPut);
  }
}
void write() { cout << ansInv << " " << ansX << "\n"; }
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0), cout.precision(20),
      cout.setf(ios::fixed);
  read();
  run();
  write();
  return 0;
}

```