#include <bits/stdc++.h>
using namespace std;
const int lg = 31;
struct node {
  long long inv, vni, mx;
  node *left, *right;
  node() : mx(0), inv(0), vni(0), left(0), right(0) {}
  ~node() {
    delete left;
    delete right;
  }
};
int n;
int A[300000];
void ins(node*& nd, int k, int i) {
  if (nd == 0) nd = new node;
  nd->mx += 1;
  if (k < 0) return;
  if (A[i] & (1 << k)) {
    ins(nd->left, k - 1, i);
    if (nd->right) nd->vni += nd->right->mx;
  } else {
    ins(nd->right, k - 1, i);
    if (nd->left) nd->inv += nd->left->mx;
  }
  return;
}
long long dp[lg + 1][2];
int mx;
void count(node* nd, int i) {
  if (!nd) return;
  count(nd->left, i - 1);
  count(nd->right, i - 1);
  dp[i][0] += nd->inv;
  dp[i][1] += nd->vni;
  return;
}
int main() {
  cin.tie(NULL)->sync_with_stdio(false);
  cin >> n;
  node* root = 0;
  for (int i = 0; i < n; ++i) {
    cin >> A[i];
    ins(root, lg, i);
  }
  count(root, lg);
  long long nx = 0;
  for (int i = lg; i > -1; --i) {
    if (dp[i][0] > dp[i][1]) {
      mx |= (1 << i);
      nx += dp[i][1];
    } else
      nx += dp[i][0];
  }
  cout << nx << ' ' << mx;
  delete root;
  return 0;
}

### Prompt

Generate 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 int lg = 31;
struct node {
  long long inv, vni, mx;
  node *left, *right;
  node() : mx(0), inv(0), vni(0), left(0), right(0) {}
  ~node() {
    delete left;
    delete right;
  }
};
int n;
int A[300000];
void ins(node*& nd, int k, int i) {
  if (nd == 0) nd = new node;
  nd->mx += 1;
  if (k < 0) return;
  if (A[i] & (1 << k)) {
    ins(nd->left, k - 1, i);
    if (nd->right) nd->vni += nd->right->mx;
  } else {
    ins(nd->right, k - 1, i);
    if (nd->left) nd->inv += nd->left->mx;
  }
  return;
}
long long dp[lg + 1][2];
int mx;
void count(node* nd, int i) {
  if (!nd) return;
  count(nd->left, i - 1);
  count(nd->right, i - 1);
  dp[i][0] += nd->inv;
  dp[i][1] += nd->vni;
  return;
}
int main() {
  cin.tie(NULL)->sync_with_stdio(false);
  cin >> n;
  node* root = 0;
  for (int i = 0; i < n; ++i) {
    cin >> A[i];
    ins(root, lg, i);
  }
  count(root, lg);
  long long nx = 0;
  for (int i = lg; i > -1; --i) {
    if (dp[i][0] > dp[i][1]) {
      mx |= (1 << i);
      nx += dp[i][1];
    } else
      nx += dp[i][0];
  }
  cout << nx << ' ' << mx;
  delete root;
  return 0;
}

```