#include <bits/stdc++.h>
#pragma GCC optimize("O2")
using namespace std;
const int N = 1e6 + 10;
const long long mod = 1e9 + 7;
const long long mod2 = 998244353;
const long long inf = 8e18;
const int LOG = 31;
long long pw(long long a, long long b, long long M) {
  return (!b ? 1
             : (b & 1 ? (a * pw(a * a % M, b / 2, M)) % M
                      : pw(a * a % M, b / 2, M)));
}
int A[N];
vector<vector<int> > vec[LOG];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  vector<int> halachikarkonam;
  for (int i = 1; i <= n; i++) {
    cin >> A[i];
    halachikarkonam.push_back(A[i]);
  }
  vec[LOG - 1].push_back(halachikarkonam);
  long long ans = 0, Xor = 0;
  for (int T = LOG - 1; ~T; T--) {
    long long sum0 = 0, sum1 = 0;
    for (auto now : vec[T]) {
      int sz = (int)now.size();
      long long cnt0 = 0, cnt1 = 0;
      for (int i = sz - 1; ~i; i--) {
        if (now[i] & (1 << T)) {
          sum0 += cnt0;
          cnt1++;
        } else {
          sum1 += cnt1;
          cnt0++;
        }
      }
    }
    ans += min(sum0, sum1);
    if (sum1 < sum0) {
      Xor |= (1 << T);
    } else {
      int wtf = 0;
    }
    if (T == 0) break;
    for (auto now : vec[T]) {
      vector<int> hala0, hala1;
      int sz = (int)now.size();
      for (int i = 0; i < sz; i++) {
        if (now[i] & (1 << T)) {
          hala1.push_back(now[i]);
        } else {
          hala0.push_back(now[i]);
        }
      }
      if (!hala0.empty()) vec[T - 1].push_back(hala0);
      if (!hala1.empty()) vec[T - 1].push_back(hala1);
    }
  }
  cout << ans << " " << Xor;
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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>
#pragma GCC optimize("O2")
using namespace std;
const int N = 1e6 + 10;
const long long mod = 1e9 + 7;
const long long mod2 = 998244353;
const long long inf = 8e18;
const int LOG = 31;
long long pw(long long a, long long b, long long M) {
  return (!b ? 1
             : (b & 1 ? (a * pw(a * a % M, b / 2, M)) % M
                      : pw(a * a % M, b / 2, M)));
}
int A[N];
vector<vector<int> > vec[LOG];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  vector<int> halachikarkonam;
  for (int i = 1; i <= n; i++) {
    cin >> A[i];
    halachikarkonam.push_back(A[i]);
  }
  vec[LOG - 1].push_back(halachikarkonam);
  long long ans = 0, Xor = 0;
  for (int T = LOG - 1; ~T; T--) {
    long long sum0 = 0, sum1 = 0;
    for (auto now : vec[T]) {
      int sz = (int)now.size();
      long long cnt0 = 0, cnt1 = 0;
      for (int i = sz - 1; ~i; i--) {
        if (now[i] & (1 << T)) {
          sum0 += cnt0;
          cnt1++;
        } else {
          sum1 += cnt1;
          cnt0++;
        }
      }
    }
    ans += min(sum0, sum1);
    if (sum1 < sum0) {
      Xor |= (1 << T);
    } else {
      int wtf = 0;
    }
    if (T == 0) break;
    for (auto now : vec[T]) {
      vector<int> hala0, hala1;
      int sz = (int)now.size();
      for (int i = 0; i < sz; i++) {
        if (now[i] & (1 << T)) {
          hala1.push_back(now[i]);
        } else {
          hala0.push_back(now[i]);
        }
      }
      if (!hala0.empty()) vec[T - 1].push_back(hala0);
      if (!hala1.empty()) vec[T - 1].push_back(hala1);
    }
  }
  cout << ans << " " << Xor;
  return 0;
}

```