#include <bits/stdc++.h>
int n, a[300000], b[300000], s[2][300000];
long long x[30], y[30];
void work(int l, int r, int lv) {
  if (lv == -1) return;
  if (l == r) return;
  int s0 = 0, s1 = 0;
  for (int i = l, k = l; i < r; i++) {
    if (a[b[i]] & (1 << lv)) {
      s[1][s1++] = b[i];
      y[lv] += s0;
    } else {
      s[0][s0++] = b[i];
      x[lv] += s1;
    }
  }
  for (int i = 0; i < s0; i++) b[l + i] = s[0][i];
  for (int i = 0; i < s1; i++) b[l + s0 + i] = s[1][i];
  work(l, l + s0, lv - 1);
  work(l + s0, r, lv - 1);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    b[i] = i;
  }
  work(0, n, 30 - 1);
  long long ans = 0;
  int t = 0;
  for (int i = 30 - 1; i >= 0; i--) {
    if (x[i] <= y[i]) {
      ans += x[i];
    } else {
      ans += y[i];
      t |= (1 << i);
    }
  }
  printf("%lld %d\n", ans, t);
}

### Prompt

Please formulate 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>
int n, a[300000], b[300000], s[2][300000];
long long x[30], y[30];
void work(int l, int r, int lv) {
  if (lv == -1) return;
  if (l == r) return;
  int s0 = 0, s1 = 0;
  for (int i = l, k = l; i < r; i++) {
    if (a[b[i]] & (1 << lv)) {
      s[1][s1++] = b[i];
      y[lv] += s0;
    } else {
      s[0][s0++] = b[i];
      x[lv] += s1;
    }
  }
  for (int i = 0; i < s0; i++) b[l + i] = s[0][i];
  for (int i = 0; i < s1; i++) b[l + s0 + i] = s[1][i];
  work(l, l + s0, lv - 1);
  work(l + s0, r, lv - 1);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    b[i] = i;
  }
  work(0, n, 30 - 1);
  long long ans = 0;
  int t = 0;
  for (int i = 30 - 1; i >= 0; i--) {
    if (x[i] <= y[i]) {
      ans += x[i];
    } else {
      ans += y[i];
      t |= (1 << i);
    }
  }
  printf("%lld %d\n", ans, t);
}

```