#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("O3")
#pragma GCC optimize("O2")
#pragma GCC optimize("Og")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
#pragma GCC optimize("-fgcse")
using namespace std;
long long MOD = 998245709;
long long exp(long long n, long long e) {
  if (e == 0) return 1;
  long long temp = exp((n * n) % MOD, e / 2);
  return e % 2 == 0 ? temp : (temp * n) % MOD;
}
long long inverse(long long n) { return exp(n, MOD - 2); }
void frac(long long p, long long q) { cout << (p * inverse(q)) % MOD << endl; }
long long ceil_div(long long a, long long b) { return (a + b - 1) / b; }
typedef struct data {
  data *bit[2];
  long long cnt = 0;
} node;
node *head;
long long inversions = 0;
vector<long long> increase(35, 0);
vector<long long> decrease(35, 0);
struct trie {
  node *head;
  void insert(long long x) {
    node *cur = head;
    cur->cnt++;
    for (long long i = 30; i >= 0; i--) {
      int a1 = cur->cnt;
      long long b = (x >> i) & 1;
      if (!cur->bit[b]) cur->bit[b] = new node();
      cur = cur->bit[b];
      cur->cnt++;
      int a2 = cur->cnt;
      int a3 = a1 - a2;
      if (b) {
        increase[i] += a3;
      } else {
        decrease[i] += a3;
        inversions += a3;
      }
    }
  }
  long long maxxor(long long x) {
    node *cur = head;
    long long ans = 0;
    for (long long i = 30; i >= 0; i--) {
      long long b = (x >> i) & 1;
      if (cur->bit[!b] && cur->bit[!b]->cnt > 0) {
        ans += (1LL << i);
        cur = cur->bit[!b];
      } else
        cur = cur->bit[b];
    }
    return ans;
  }
};
trie root;
void solve() {
  int n;
  cin >> n;
  root.head = new node();
  vector<long long> arr(n + 1);
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
    root.insert(arr[i]);
  }
  long long x = 0;
  for (int i = 0; i <= 30; i++) {
    if (increase[i] < decrease[i]) {
      x += (1LL << i);
      inversions -= decrease[i];
      inversions += increase[i];
    }
  }
  cout << inversions << " " << x << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  for (long long i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("O3")
#pragma GCC optimize("O2")
#pragma GCC optimize("Og")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
#pragma GCC optimize("-fgcse")
using namespace std;
long long MOD = 998245709;
long long exp(long long n, long long e) {
  if (e == 0) return 1;
  long long temp = exp((n * n) % MOD, e / 2);
  return e % 2 == 0 ? temp : (temp * n) % MOD;
}
long long inverse(long long n) { return exp(n, MOD - 2); }
void frac(long long p, long long q) { cout << (p * inverse(q)) % MOD << endl; }
long long ceil_div(long long a, long long b) { return (a + b - 1) / b; }
typedef struct data {
  data *bit[2];
  long long cnt = 0;
} node;
node *head;
long long inversions = 0;
vector<long long> increase(35, 0);
vector<long long> decrease(35, 0);
struct trie {
  node *head;
  void insert(long long x) {
    node *cur = head;
    cur->cnt++;
    for (long long i = 30; i >= 0; i--) {
      int a1 = cur->cnt;
      long long b = (x >> i) & 1;
      if (!cur->bit[b]) cur->bit[b] = new node();
      cur = cur->bit[b];
      cur->cnt++;
      int a2 = cur->cnt;
      int a3 = a1 - a2;
      if (b) {
        increase[i] += a3;
      } else {
        decrease[i] += a3;
        inversions += a3;
      }
    }
  }
  long long maxxor(long long x) {
    node *cur = head;
    long long ans = 0;
    for (long long i = 30; i >= 0; i--) {
      long long b = (x >> i) & 1;
      if (cur->bit[!b] && cur->bit[!b]->cnt > 0) {
        ans += (1LL << i);
        cur = cur->bit[!b];
      } else
        cur = cur->bit[b];
    }
    return ans;
  }
};
trie root;
void solve() {
  int n;
  cin >> n;
  root.head = new node();
  vector<long long> arr(n + 1);
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
    root.insert(arr[i]);
  }
  long long x = 0;
  for (int i = 0; i <= 30; i++) {
    if (increase[i] < decrease[i]) {
      x += (1LL << i);
      inversions -= decrease[i];
      inversions += increase[i];
    }
  }
  cout << inversions << " " << x << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  for (long long i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}

```