#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5;
const int M = 4e6 + 5;
int n;
int a[N];
int tire[M][2];
long long dp[32][2];
long long ans1, ans2;
int tot;
vector<int> v[M];
void insert(int x, int pos) {
  int rt = 0, t;
  for (int i = 30; i >= 0; i--) {
    t = (x >> i) & 1;
    if (!tire[rt][t]) tire[rt][t] = ++tot;
    rt = tire[rt][t];
    v[rt].push_back(pos);
  }
}
void dfs(int rt, int pos) {
  if (pos == -1) return;
  int l = tire[rt][0];
  int r = tire[rt][1];
  long long sz0 = v[l].size();
  long long sz1 = v[r].size();
  long long res = 0;
  long long tmp = 0;
  for (int i = 0; i < sz0; ++i) {
    while (tmp < sz1 && v[r][tmp] < v[l][i]) tmp++;
    res += tmp;
  }
  dp[pos][0] += res;
  dp[pos][1] += sz0 * sz1 - res;
  if (l != 0) dfs(l, pos - 1);
  if (r != 0) dfs(r, pos - 1);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    insert(a[i], i);
  }
  dfs(0, 30);
  for (int i = 0; i <= 30; ++i) {
    if (dp[i][0] <= dp[i][1]) {
      ans1 += dp[i][0];
    } else {
      ans1 += dp[i][1];
      ans2 += (1 << i);
    }
  }
  printf("%lld %lld\n", ans1, ans2);
  return 0;
}

### 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;
const int N = 3e5 + 5;
const int M = 4e6 + 5;
int n;
int a[N];
int tire[M][2];
long long dp[32][2];
long long ans1, ans2;
int tot;
vector<int> v[M];
void insert(int x, int pos) {
  int rt = 0, t;
  for (int i = 30; i >= 0; i--) {
    t = (x >> i) & 1;
    if (!tire[rt][t]) tire[rt][t] = ++tot;
    rt = tire[rt][t];
    v[rt].push_back(pos);
  }
}
void dfs(int rt, int pos) {
  if (pos == -1) return;
  int l = tire[rt][0];
  int r = tire[rt][1];
  long long sz0 = v[l].size();
  long long sz1 = v[r].size();
  long long res = 0;
  long long tmp = 0;
  for (int i = 0; i < sz0; ++i) {
    while (tmp < sz1 && v[r][tmp] < v[l][i]) tmp++;
    res += tmp;
  }
  dp[pos][0] += res;
  dp[pos][1] += sz0 * sz1 - res;
  if (l != 0) dfs(l, pos - 1);
  if (r != 0) dfs(r, pos - 1);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    insert(a[i], i);
  }
  dfs(0, 30);
  for (int i = 0; i <= 30; ++i) {
    if (dp[i][0] <= dp[i][1]) {
      ans1 += dp[i][0];
    } else {
      ans1 += dp[i][1];
      ans2 += (1 << i);
    }
  }
  printf("%lld %lld\n", ans1, ans2);
  return 0;
}

```