#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 10;
const int INF = 0x3f3f3f3f;
const long long inf = 0x3f3f3f3f3f3f3f3f;
const long long mod = 1e9 + 7;
int n;
int a[N];
vector<int> vc;
int m;
long long ans, x, co[100][2];
void dfs(int dep, vector<int>& vc) {
  if (dep < 0 || vc.empty()) {
    return;
  }
  vector<int> t[2];
  long long cnt0 = 0, cnt1 = 0;
  for (int u : vc) {
    if ((u >> dep & 1) == 0)
      cnt0 += (int)t[1].size();
    else
      cnt1 += (int)t[0].size();
    t[u >> dep & 1].push_back(u);
  }
  co[dep][0] += cnt0;
  co[dep][1] += cnt1;
  dfs(dep - 1, t[0]);
  dfs(dep - 1, t[1]);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    vc.push_back(a[i]);
    int x = a[i], c = 0;
    while (x) {
      c++;
      x >>= 1;
    }
    m = max(m, c);
  }
  dfs(m - 1, vc);
  for (int i = m - 1; i >= 0; i--) {
    if (co[i][0] > co[i][1]) x |= (1ll << i);
    ans += min(co[i][0], co[i][1]);
  }
  printf("%lld %lld\n", ans, x);
  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>
using namespace std;
const int N = 2e6 + 10;
const int INF = 0x3f3f3f3f;
const long long inf = 0x3f3f3f3f3f3f3f3f;
const long long mod = 1e9 + 7;
int n;
int a[N];
vector<int> vc;
int m;
long long ans, x, co[100][2];
void dfs(int dep, vector<int>& vc) {
  if (dep < 0 || vc.empty()) {
    return;
  }
  vector<int> t[2];
  long long cnt0 = 0, cnt1 = 0;
  for (int u : vc) {
    if ((u >> dep & 1) == 0)
      cnt0 += (int)t[1].size();
    else
      cnt1 += (int)t[0].size();
    t[u >> dep & 1].push_back(u);
  }
  co[dep][0] += cnt0;
  co[dep][1] += cnt1;
  dfs(dep - 1, t[0]);
  dfs(dep - 1, t[1]);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    vc.push_back(a[i]);
    int x = a[i], c = 0;
    while (x) {
      c++;
      x >>= 1;
    }
    m = max(m, c);
  }
  dfs(m - 1, vc);
  for (int i = m - 1; i >= 0; i--) {
    if (co[i][0] > co[i][1]) x |= (1ll << i);
    ans += min(co[i][0], co[i][1]);
  }
  printf("%lld %lld\n", ans, x);
  return 0;
}

```