#include <bits/stdc++.h>
using namespace std;
long long int n, m;
long long int arr[300000];
vector<long long int> adj[6000500];
long long int child[2][6000500];
long long int dp[30][2];
long long int ind = 1;
void add(long long int num, long long int pos) {
  long long int cur = 0;
  for (long long int i = 29; i >= 0; i--) {
    long long int bit = ((num >> i) & 1);
    if (!child[bit][cur]) child[bit][cur] = ind++;
    adj[child[bit][cur]].push_back(pos);
    cur = child[bit][cur];
  }
}
void go(long long int node, long long int lev) {
  long long int l = child[0][node];
  long long int r = child[1][node];
  if (l) go(l, lev - 1);
  if (r) go(r, lev - 1);
  if (!l || !r) return;
  long long int ret = 0;
  long long int j = 0;
  for (auto it : adj[l]) {
    while (j < adj[r].size() && it > adj[r][j]) j++;
    ret += j;
  }
  dp[lev][0] += ret;
  dp[lev][1] += (adj[r].size() * 1ll * adj[l].size()) - ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int TESTS, q, a, b, l, r, c, k, p, h, w, x, y, z, xs, ys, t, f;
  long double d;
  TESTS = 1;
  memset(dp, 0, sizeof(dp));
  memset(child, 0, sizeof(child));
  while (TESTS--) {
    cin >> n;
    long long int maxi = 0;
    for (long long int i = 0; i <= n - 1; i++) {
      cin >> arr[i];
      add(arr[i], i);
    }
    go(0, 29);
    b = 0;
    x = 0;
    long long int cur = 1;
    for (long long int i = 0; i <= 29; i++) {
      if (dp[i][1] < dp[i][0]) x += cur;
      b += min(dp[i][1], dp[i][0]);
      cur *= 2;
    }
    cout << b << " " << x;
  }
  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;
long long int n, m;
long long int arr[300000];
vector<long long int> adj[6000500];
long long int child[2][6000500];
long long int dp[30][2];
long long int ind = 1;
void add(long long int num, long long int pos) {
  long long int cur = 0;
  for (long long int i = 29; i >= 0; i--) {
    long long int bit = ((num >> i) & 1);
    if (!child[bit][cur]) child[bit][cur] = ind++;
    adj[child[bit][cur]].push_back(pos);
    cur = child[bit][cur];
  }
}
void go(long long int node, long long int lev) {
  long long int l = child[0][node];
  long long int r = child[1][node];
  if (l) go(l, lev - 1);
  if (r) go(r, lev - 1);
  if (!l || !r) return;
  long long int ret = 0;
  long long int j = 0;
  for (auto it : adj[l]) {
    while (j < adj[r].size() && it > adj[r][j]) j++;
    ret += j;
  }
  dp[lev][0] += ret;
  dp[lev][1] += (adj[r].size() * 1ll * adj[l].size()) - ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int TESTS, q, a, b, l, r, c, k, p, h, w, x, y, z, xs, ys, t, f;
  long double d;
  TESTS = 1;
  memset(dp, 0, sizeof(dp));
  memset(child, 0, sizeof(child));
  while (TESTS--) {
    cin >> n;
    long long int maxi = 0;
    for (long long int i = 0; i <= n - 1; i++) {
      cin >> arr[i];
      add(arr[i], i);
    }
    go(0, 29);
    b = 0;
    x = 0;
    long long int cur = 1;
    for (long long int i = 0; i <= 29; i++) {
      if (dp[i][1] < dp[i][0]) x += cur;
      b += min(dp[i][1], dp[i][0]);
      cur *= 2;
    }
    cout << b << " " << x;
  }
  return 0;
}

```