#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int N, A[300005], B[300005], C[300005];
ll mergesort(int s, int e) {
  if (s == e) return 0;
  int mid = (s + e) >> 1;
  ll res = mergesort(s, mid) + mergesort(mid + 1, e);
  for (int i = s; i <= e; i++) C[i] = B[i];
  int i1 = s, i2 = mid + 1;
  for (int x = 0; x <= e - s; x++) {
    if (i1 > mid) {
      B[s + x] = C[i2++];
    } else if (i2 > e) {
      B[s + x] = C[i1++];
      res += i2 - mid - 1;
    } else if (C[i1] <= C[i2]) {
      B[s + x] = C[i1++];
      res += i2 - mid - 1;
    } else {
      B[s + x] = C[i2++];
    }
  }
  return res;
}
ll go(int bit) {
  for (int i = 1; i <= N; i++) B[i] = (A[i] ^ bit);
  return mergesort(1, N);
}
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%d", A + i);
  int ans = 0;
  ll cur = go(0);
  for (int i = 30; ~i; i--) {
    ll res = go(ans | (1 << i));
    if (cur > res) {
      cur = res;
      ans |= (1 << i);
    }
  }
  printf("%lld %d\n", cur, ans);
}

### Prompt

In cpp, your task is to solve 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;
using ll = long long;
int N, A[300005], B[300005], C[300005];
ll mergesort(int s, int e) {
  if (s == e) return 0;
  int mid = (s + e) >> 1;
  ll res = mergesort(s, mid) + mergesort(mid + 1, e);
  for (int i = s; i <= e; i++) C[i] = B[i];
  int i1 = s, i2 = mid + 1;
  for (int x = 0; x <= e - s; x++) {
    if (i1 > mid) {
      B[s + x] = C[i2++];
    } else if (i2 > e) {
      B[s + x] = C[i1++];
      res += i2 - mid - 1;
    } else if (C[i1] <= C[i2]) {
      B[s + x] = C[i1++];
      res += i2 - mid - 1;
    } else {
      B[s + x] = C[i2++];
    }
  }
  return res;
}
ll go(int bit) {
  for (int i = 1; i <= N; i++) B[i] = (A[i] ^ bit);
  return mergesort(1, N);
}
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) scanf("%d", A + i);
  int ans = 0;
  ll cur = go(0);
  for (int i = 30; ~i; i--) {
    ll res = go(ans | (1 << i));
    if (cur > res) {
      cur = res;
      ans |= (1 << i);
    }
  }
  printf("%lld %d\n", cur, ans);
}

```