#include <bits/stdc++.h>
using namespace std;
const long long mask = (1ll << 31) - 1;
const int N = 3e5 + 10;
int a[N], b[N], c[N], d[N], id[N];
vector<int> O, Z;
void add(int u, int d) {
  for (; u < N; u += u & -u) b[u] += d;
}
int calc(int u, int ans = 0) {
  for (; u > 0; u -= u & -u) ans += b[u];
  return ans;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
    id[i] = i;
  }
  long long inv = 0;
  int ans = 0;
  for (int i = 0; i < 31; i++) {
    for (int j = 1; j <= n; j++) {
      c[j] = a[j] & (mask << i + 1);
      d[j] = a[j] - c[j];
    }
    sort(id + 1, id + n + 1, [&](int x, int y) {
      if (c[x] != c[y]) return c[x] < c[y];
      return d[x] < d[y];
    });
    long long tot = 0, tmp = 0;
    for (int j = 1, k; j <= n; j = k) {
      O.clear();
      Z.clear();
      int zero = 0, one = 0;
      for (k = j; k <= n && c[id[j]] == c[id[k]]; k++) {
        if (d[id[k]] & (1 << i))
          one++, O.push_back(id[k]);
        else
          zero++, Z.push_back(id[k]);
      }
      for (auto r : O) add(r, 1);
      for (auto r : Z) tmp += calc(r);
      for (auto r : O) add(r, -1);
      tot += 1ll * zero * one;
    }
    if (tot - tmp < tmp) {
      ans ^= 1 << i;
    }
    inv += min(tot - tmp, tmp);
  }
  cout << inv << ' ' << ans << endl;
  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>
using namespace std;
const long long mask = (1ll << 31) - 1;
const int N = 3e5 + 10;
int a[N], b[N], c[N], d[N], id[N];
vector<int> O, Z;
void add(int u, int d) {
  for (; u < N; u += u & -u) b[u] += d;
}
int calc(int u, int ans = 0) {
  for (; u > 0; u -= u & -u) ans += b[u];
  return ans;
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
    id[i] = i;
  }
  long long inv = 0;
  int ans = 0;
  for (int i = 0; i < 31; i++) {
    for (int j = 1; j <= n; j++) {
      c[j] = a[j] & (mask << i + 1);
      d[j] = a[j] - c[j];
    }
    sort(id + 1, id + n + 1, [&](int x, int y) {
      if (c[x] != c[y]) return c[x] < c[y];
      return d[x] < d[y];
    });
    long long tot = 0, tmp = 0;
    for (int j = 1, k; j <= n; j = k) {
      O.clear();
      Z.clear();
      int zero = 0, one = 0;
      for (k = j; k <= n && c[id[j]] == c[id[k]]; k++) {
        if (d[id[k]] & (1 << i))
          one++, O.push_back(id[k]);
        else
          zero++, Z.push_back(id[k]);
      }
      for (auto r : O) add(r, 1);
      for (auto r : Z) tmp += calc(r);
      for (auto r : O) add(r, -1);
      tot += 1ll * zero * one;
    }
    if (tot - tmp < tmp) {
      ans ^= 1 << i;
    }
    inv += min(tot - tmp, tmp);
  }
  cout << inv << ' ' << ans << endl;
  return 0;
}

```