#include <bits/stdc++.h>
using namespace std;
long long int pow_(long long int a, long long int e,
                   long long int m = 1000000007) {
  long long int r = 1;
  while (e) {
    if (e & 1) r = r * a % m;
    a = a * a % m;
    e >>= 1;
  }
  return r;
}
int inv(long long int a) { return pow_(a, 1000000007 - 2, 1000000007); }
int add(long long int a, long long int b) {
  return (a + b + 1000000007) % 1000000007;
}
int mul(long long int a, long long int b) {
  return (1000000007 + a * b % 1000000007) % 1000000007;
}
int di(long long int a, long long int b) { return mul(a, inv(b)); }
long long int ceil(long long int a, long long int b) {
  return a / b + (a % b != 0);
}
void print(vector<int> a) {
  for (int i = 0; i < a.size(); i++) cout << a[i] << " ";
  cout << '\n';
}
void print(long long int a, long long int b, char k = '\n') {
  cout << a << " " << b << k;
}
void print(vector<int> *a, int n) {
  for (int i = 0; i < n; i++) print(a[i]);
}
void print(int *a, int n) {
  for (int i = 0; i < n; i++) cout << a[i] << " ";
  cout << '\n';
}
long long int calc[2][1000005];
int N;
void find(vector<int> &A, int l) {
  if (A.size() < 2) return;
  if (l < 0) return;
  vector<int> X, Y;
  int Z = 1 << l;
  int c0 = 0, c1 = 0;
  long long int &in0 = calc[0][l], &in1 = calc[1][l];
  for (auto i : A) {
    if (i & Z) {
      c1 += 1;
      in1 += c0;
      X.push_back(i);
    } else {
      c0 += 1;
      in0 += c1;
      Y.push_back(i);
    }
  }
  find(X, l - 1);
  find(Y, l - 1);
}
int solve() {
  cin >> N;
  vector<int> A(N);
  int l = 0;
  for (auto &i : A) {
    cin >> i;
    l = max(l, (int)log2(i) + 1);
  }
  find(A, l);
  long long int v = 0, c = 0;
  for (int i = l; - 1 > 0 ? i < -1 : i > -1; i += -1) {
    v <<= 1;
    c += min(calc[0][i], calc[1][i]);
    if (calc[1][i] < calc[0][i]) v |= 1;
  }
  cout << c << " " << v;
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
  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;
long long int pow_(long long int a, long long int e,
                   long long int m = 1000000007) {
  long long int r = 1;
  while (e) {
    if (e & 1) r = r * a % m;
    a = a * a % m;
    e >>= 1;
  }
  return r;
}
int inv(long long int a) { return pow_(a, 1000000007 - 2, 1000000007); }
int add(long long int a, long long int b) {
  return (a + b + 1000000007) % 1000000007;
}
int mul(long long int a, long long int b) {
  return (1000000007 + a * b % 1000000007) % 1000000007;
}
int di(long long int a, long long int b) { return mul(a, inv(b)); }
long long int ceil(long long int a, long long int b) {
  return a / b + (a % b != 0);
}
void print(vector<int> a) {
  for (int i = 0; i < a.size(); i++) cout << a[i] << " ";
  cout << '\n';
}
void print(long long int a, long long int b, char k = '\n') {
  cout << a << " " << b << k;
}
void print(vector<int> *a, int n) {
  for (int i = 0; i < n; i++) print(a[i]);
}
void print(int *a, int n) {
  for (int i = 0; i < n; i++) cout << a[i] << " ";
  cout << '\n';
}
long long int calc[2][1000005];
int N;
void find(vector<int> &A, int l) {
  if (A.size() < 2) return;
  if (l < 0) return;
  vector<int> X, Y;
  int Z = 1 << l;
  int c0 = 0, c1 = 0;
  long long int &in0 = calc[0][l], &in1 = calc[1][l];
  for (auto i : A) {
    if (i & Z) {
      c1 += 1;
      in1 += c0;
      X.push_back(i);
    } else {
      c0 += 1;
      in0 += c1;
      Y.push_back(i);
    }
  }
  find(X, l - 1);
  find(Y, l - 1);
}
int solve() {
  cin >> N;
  vector<int> A(N);
  int l = 0;
  for (auto &i : A) {
    cin >> i;
    l = max(l, (int)log2(i) + 1);
  }
  find(A, l);
  long long int v = 0, c = 0;
  for (int i = l; - 1 > 0 ? i < -1 : i > -1; i += -1) {
    v <<= 1;
    c += min(calc[0][i], calc[1][i]);
    if (calc[1][i] < calc[0][i]) v |= 1;
  }
  cout << c << " " << v;
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```