#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
const int MOD = 1e9 + 7;
const int dx[] = {0, 0, 1, -1};
const int dy[] = {1, -1, 0, 0};
const char dir[] = {'R', 'L', 'D', 'U'};
ll add(ll a, ll b) {
  a += b;
  if (a >= MOD) {
    a -= MOD;
  }
  return a;
}
int sub(int a, int b) {
  a -= b;
  if (a < 0) a += MOD;
  return a;
}
int mult(int a, int b) { return ((ll)a * b) % MOD; }
void setIO() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
}
int n;
vector<vector<int>> sets[30];
int main() {
  setIO();
  cin >> n;
  int a[n];
  vector<int> orig;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    orig.emplace_back(i);
  }
  sets[29].emplace_back(orig);
  int first = 0;
  ll inv = 0;
  for (int bt = 29; bt >= 0; --bt) {
    ll tot01 = 0, tot10 = 0;
    for (vector<int> i : sets[bt]) {
      int cnt[2] = {0, 0};
      ll inv01 = 0;
      ll inv10 = 0;
      vector<int> zero, one;
      for (int j : i) {
        if (a[j] & (1 << bt)) {
          ++cnt[1];
          inv10 += cnt[0];
          one.emplace_back(j);
        } else {
          ++cnt[0];
          inv01 += cnt[1];
          zero.emplace_back(j);
        }
      }
      tot01 += inv01;
      tot10 += inv10;
      if (bt) {
        if (zero.size()) sets[bt - 1].emplace_back(zero);
        if (one.size()) sets[bt - 1].emplace_back(one);
      }
    }
    if (tot01 > tot10) {
      first += (1 << bt);
      inv += tot10;
    } else
      inv += tot01;
  }
  cout << inv << " " << first << "\n";
}

### Prompt

Generate 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;
using ll = long long;
using pii = pair<int, int>;
const int MOD = 1e9 + 7;
const int dx[] = {0, 0, 1, -1};
const int dy[] = {1, -1, 0, 0};
const char dir[] = {'R', 'L', 'D', 'U'};
ll add(ll a, ll b) {
  a += b;
  if (a >= MOD) {
    a -= MOD;
  }
  return a;
}
int sub(int a, int b) {
  a -= b;
  if (a < 0) a += MOD;
  return a;
}
int mult(int a, int b) { return ((ll)a * b) % MOD; }
void setIO() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
}
int n;
vector<vector<int>> sets[30];
int main() {
  setIO();
  cin >> n;
  int a[n];
  vector<int> orig;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    orig.emplace_back(i);
  }
  sets[29].emplace_back(orig);
  int first = 0;
  ll inv = 0;
  for (int bt = 29; bt >= 0; --bt) {
    ll tot01 = 0, tot10 = 0;
    for (vector<int> i : sets[bt]) {
      int cnt[2] = {0, 0};
      ll inv01 = 0;
      ll inv10 = 0;
      vector<int> zero, one;
      for (int j : i) {
        if (a[j] & (1 << bt)) {
          ++cnt[1];
          inv10 += cnt[0];
          one.emplace_back(j);
        } else {
          ++cnt[0];
          inv01 += cnt[1];
          zero.emplace_back(j);
        }
      }
      tot01 += inv01;
      tot10 += inv10;
      if (bt) {
        if (zero.size()) sets[bt - 1].emplace_back(zero);
        if (one.size()) sets[bt - 1].emplace_back(one);
      }
    }
    if (tot01 > tot10) {
      first += (1 << bt);
      inv += tot10;
    } else
      inv += tot01;
  }
  cout << inv << " " << first << "\n";
}

```