#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast")
struct chash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
int main() {
  long long sum = 0;
  int n, a, m, k, b, c, d, p, r, i, Ans = 0, mask = 0;
  cin >> n;
  vector<int> A(n);
  for (int i = 0; i < n; i++) scanf("%i", &A[i]);
  for (int i = 1 << 29; i; i >>= 1) {
    mask |= i;
    long long q10 = 0, q01 = 0;
    static unordered_map<int, int> M;
    M.reserve(3e5);
    for (int j = 0; j < n; j++) {
      M[A[j] & mask]++;
      ((A[j] & i) ? q01 : q10) += M[(A[j] ^ i) & mask];
    }
    M.clear();
    if (q10 > q01) Ans |= i;
    sum += min(q10, q01);
  }
  cout << sum << " " << Ans;
  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;
#pragma GCC optimize("Ofast")
struct chash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
int main() {
  long long sum = 0;
  int n, a, m, k, b, c, d, p, r, i, Ans = 0, mask = 0;
  cin >> n;
  vector<int> A(n);
  for (int i = 0; i < n; i++) scanf("%i", &A[i]);
  for (int i = 1 << 29; i; i >>= 1) {
    mask |= i;
    long long q10 = 0, q01 = 0;
    static unordered_map<int, int> M;
    M.reserve(3e5);
    for (int j = 0; j < n; j++) {
      M[A[j] & mask]++;
      ((A[j] & i) ? q01 : q10) += M[(A[j] ^ i) & mask];
    }
    M.clear();
    if (q10 > q01) Ans |= i;
    sum += min(q10, q01);
  }
  cout << sum << " " << Ans;
  return 0;
}

```