#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const int MAXN = 3e5 + 10;
int nxt[30];
long long dp[30];
int dat[MAXN];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &dat[i]);
  }
  for (int b = 0; b < 30; b++) {
    long long op0 = 0;
    long long op1 = 0;
    unordered_map<int, long long> mapa;
    mapa.max_load_factor(0.25);
    for (int i = 0; i < n; i++) {
      long long cant = mapa[dat[i] ^ 1];
      if (dat[i] & 1) {
        op1 += cant;
      } else {
        op0 += cant;
      }
      mapa[dat[i]]++;
      dat[i] >>= 1;
    }
    if (op0 <= op1)
      nxt[b] = 0;
    else
      nxt[b] = 1;
    dp[b] = min(op0, op1);
    if (b > 0) dp[b] += dp[b - 1];
  }
  int x = 0;
  for (int b = 0; b < 30; b++) {
    if (nxt[b]) x += (1 << b);
  }
  printf("%lld %d\n", dp[29], x);
  return 0;
}

### Prompt

Your challenge is to write 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>
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const int MAXN = 3e5 + 10;
int nxt[30];
long long dp[30];
int dat[MAXN];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &dat[i]);
  }
  for (int b = 0; b < 30; b++) {
    long long op0 = 0;
    long long op1 = 0;
    unordered_map<int, long long> mapa;
    mapa.max_load_factor(0.25);
    for (int i = 0; i < n; i++) {
      long long cant = mapa[dat[i] ^ 1];
      if (dat[i] & 1) {
        op1 += cant;
      } else {
        op0 += cant;
      }
      mapa[dat[i]]++;
      dat[i] >>= 1;
    }
    if (op0 <= op1)
      nxt[b] = 0;
    else
      nxt[b] = 1;
    dp[b] = min(op0, op1);
    if (b > 0) dp[b] += dp[b - 1];
  }
  int x = 0;
  for (int b = 0; b < 30; b++) {
    if (nxt[b]) x += (1 << b);
  }
  printf("%lld %d\n", dp[29], x);
  return 0;
}

```