#include <bits/stdc++.h>
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
const int MOD = 1e9 + 7;
const int INF = 1e9 + 9;
const int MX = 3e5 + 5;
long long invCnt[32][2];
void solve(vector<long long> inp, long long idx) {
  if (inp.size() <= 1) {
    return;
  }
  if (idx < 0) {
    return;
  }
  std::vector<long long> t0, t1;
  for (long long i = 0; i < inp.size(); i++) {
    if (inp[i] & (1 << idx)) {
      t1.push_back(inp[i]);
    } else {
      t0.push_back(inp[i]);
    }
  }
  long long inv0 = 0, inv1 = 0;
  long long cnt0 = 0, cnt1 = 0;
  for (long long i = 0; i < inp.size(); i++) {
    if (inp[i] & (1 << idx)) {
      cnt1++;
      inv1 += cnt0;
    } else {
      cnt0++;
      inv0 += cnt1;
    }
  }
  invCnt[idx][0] += inv0;
  invCnt[idx][1] += inv1;
  solve(t0, idx - 1);
  solve(t1, idx - 1);
}
int main() {
  long long n;
  cin >> n;
  std::vector<long long> inp(n);
  for (long long i = 0; i < n; i++) {
    cin >> inp[i];
  }
  solve(inp, 31);
  long long ans = 0;
  long long totInv = 0;
  for (long long i = 31; i >= 0; i--) {
    long long b;
    if (invCnt[i][0] <= invCnt[i][1]) {
      b = 0;
    } else {
      b = 1;
    }
    totInv += invCnt[i][b];
    ans = ans * 2 + b;
  }
  cout << totInv << " " << ans << endl;
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
const int MOD = 1e9 + 7;
const int INF = 1e9 + 9;
const int MX = 3e5 + 5;
long long invCnt[32][2];
void solve(vector<long long> inp, long long idx) {
  if (inp.size() <= 1) {
    return;
  }
  if (idx < 0) {
    return;
  }
  std::vector<long long> t0, t1;
  for (long long i = 0; i < inp.size(); i++) {
    if (inp[i] & (1 << idx)) {
      t1.push_back(inp[i]);
    } else {
      t0.push_back(inp[i]);
    }
  }
  long long inv0 = 0, inv1 = 0;
  long long cnt0 = 0, cnt1 = 0;
  for (long long i = 0; i < inp.size(); i++) {
    if (inp[i] & (1 << idx)) {
      cnt1++;
      inv1 += cnt0;
    } else {
      cnt0++;
      inv0 += cnt1;
    }
  }
  invCnt[idx][0] += inv0;
  invCnt[idx][1] += inv1;
  solve(t0, idx - 1);
  solve(t1, idx - 1);
}
int main() {
  long long n;
  cin >> n;
  std::vector<long long> inp(n);
  for (long long i = 0; i < n; i++) {
    cin >> inp[i];
  }
  solve(inp, 31);
  long long ans = 0;
  long long totInv = 0;
  for (long long i = 31; i >= 0; i--) {
    long long b;
    if (invCnt[i][0] <= invCnt[i][1]) {
      b = 0;
    } else {
      b = 1;
    }
    totInv += invCnt[i][b];
    ans = ans * 2 + b;
  }
  cout << totInv << " " << ans << endl;
  return 0;
}

```