#include <bits/stdc++.h>
using namespace std;
int n, id;
long long dp[2][30];
int t[2][5000006];
vector<int> adj[5000006];
void add_edge(int x, int pos) {
  int v = 0;
  for (int i = 29; i >= 0; i--) {
    int bit = ((x >> i) & 1);
    if (!t[bit][v]) t[bit][v] = ++id;
    v = t[bit][v];
    adj[v].push_back(pos);
  }
}
void rec(int v, int b = 29) {
  int l = t[0][v], r = t[1][v];
  if (l) rec(l, b - 1);
  if (r) rec(r, b - 1);
  if (!l || !r) return;
  long long res = 0;
  int cur = 0;
  for (int u : adj[l]) {
    while (cur < adj[r].size() && adj[r][cur] < u) cur++;
    res += cur;
  }
  dp[0][b] += res;
  dp[1][b] += 1LL * (int)(adj[l]).size() * (int)(adj[r]).size() - res;
}
signed main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    add_edge(x, i);
  }
  rec(0);
  long long cnt = 0;
  int res = 0;
  for (int i = 0; i < 30; i++) {
    cnt += min(dp[0][i], dp[1][i]);
    if (dp[1][i] < dp[0][i]) {
      res += (1 << i);
    }
  }
  cout << cnt << ' ' << res << '\n';
  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;
int n, id;
long long dp[2][30];
int t[2][5000006];
vector<int> adj[5000006];
void add_edge(int x, int pos) {
  int v = 0;
  for (int i = 29; i >= 0; i--) {
    int bit = ((x >> i) & 1);
    if (!t[bit][v]) t[bit][v] = ++id;
    v = t[bit][v];
    adj[v].push_back(pos);
  }
}
void rec(int v, int b = 29) {
  int l = t[0][v], r = t[1][v];
  if (l) rec(l, b - 1);
  if (r) rec(r, b - 1);
  if (!l || !r) return;
  long long res = 0;
  int cur = 0;
  for (int u : adj[l]) {
    while (cur < adj[r].size() && adj[r][cur] < u) cur++;
    res += cur;
  }
  dp[0][b] += res;
  dp[1][b] += 1LL * (int)(adj[l]).size() * (int)(adj[r]).size() - res;
}
signed main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    add_edge(x, i);
  }
  rec(0);
  long long cnt = 0;
  int res = 0;
  for (int i = 0; i < 30; i++) {
    cnt += min(dp[0][i], dp[1][i]);
    if (dp[1][i] < dp[0][i]) {
      res += (1 << i);
    }
  }
  cout << cnt << ' ' << res << '\n';
  return 0;
}

```