#include <bits/stdc++.h>
using namespace std;
long long MOD = 998244353;
vector<long long> fact(3e5 + 5, 1);
void fill_fact() {
  for (int i = 2; i < fact.size(); i++) fact[i] = (fact[i - 1] * i) % MOD;
}
long long pow(long long a, long long b) {
  if (b == 0) {
    return 1;
  } else if (b % 2 == 0) {
    long long x = pow(a, b / 2);
    return x * x;
  } else {
    return a * pow(a, b - 1);
  }
}
long long pow_mod(long long a, long long b) {
  if (b == 0) {
    return 1;
  } else if (b % 2 == 0) {
    long long x = pow(a, b / 2) % MOD;
    return (x * x) % MOD;
  } else {
    return (a * pow(a, b - 1)) % MOD;
  }
}
long long cnk(long long n, long long k) {
  long long n_fact = fact[n];
  long long k_fact = fact[k];
  long long nk_fact = fact[n - k];
  long long down = (k_fact * nk_fact) % MOD;
  long long up = pow(down, MOD - 2);
  return (n_fact * up) % MOD;
}
long long answer = 0;
long long total_inv = 0;
long long n;
vector<pair<long long, long long>> LEVELS(33, make_pair(0, 0));
inline void print(const vector<int>& bits) {
  long long inv = 0, ones = 0;
  cerr << "BITS: ";
  for (int i = 0; i < bits.size(); i++) {
    cerr << bits[i];
  }
  cerr << endl;
}
inline long long inv_count(const vector<int>& bits) {
  long long inv = 0, ones = 0;
  for (int i = 0; i < bits.size(); i++) {
    if (bits[i] == 1)
      ones++;
    else
      inv += ones;
  }
  return inv;
}
inline long long inv_count2(const vector<int>& bits) {
  long long inv = 0, zeros = 0;
  for (int i = 0; i < bits.size(); i++) {
    if (bits[i] == 0)
      zeros++;
    else
      inv += zeros;
  }
  return inv;
}
void dfs(vector<long long> vec, long long level) {
  vector<int> bits;
  vector<long long> ones, zeros;
  long long BIT = (1 << level);
  for (int i = 0; i < vec.size(); i++) {
    if (vec[i] & BIT) {
      bits.push_back(1);
      ones.push_back(vec[i]);
    } else {
      bits.push_back(0);
      zeros.push_back(vec[i]);
    }
  }
  long long q = inv_count(bits);
  long long q1 = inv_count2(bits);
  LEVELS[level].first += q;
  LEVELS[level].second += q1;
  if (level > 0) {
    if (!ones.empty()) dfs(ones, level - 1);
    if (!zeros.empty()) dfs(zeros, level - 1);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  vector<long long> vec(n);
  for (int i = 0; i < n; i++) {
    cin >> vec[i];
  }
  dfs(vec, 31);
  for (long long level = 0; level <= 31; level++) {
    long long BIT = (1 << level);
    if (LEVELS[level].second < LEVELS[level].first) {
      answer |= BIT;
      total_inv += LEVELS[level].second;
    } else {
      total_inv += LEVELS[level].first;
    }
  }
  cout << total_inv << " " << answer << endl;
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 MOD = 998244353;
vector<long long> fact(3e5 + 5, 1);
void fill_fact() {
  for (int i = 2; i < fact.size(); i++) fact[i] = (fact[i - 1] * i) % MOD;
}
long long pow(long long a, long long b) {
  if (b == 0) {
    return 1;
  } else if (b % 2 == 0) {
    long long x = pow(a, b / 2);
    return x * x;
  } else {
    return a * pow(a, b - 1);
  }
}
long long pow_mod(long long a, long long b) {
  if (b == 0) {
    return 1;
  } else if (b % 2 == 0) {
    long long x = pow(a, b / 2) % MOD;
    return (x * x) % MOD;
  } else {
    return (a * pow(a, b - 1)) % MOD;
  }
}
long long cnk(long long n, long long k) {
  long long n_fact = fact[n];
  long long k_fact = fact[k];
  long long nk_fact = fact[n - k];
  long long down = (k_fact * nk_fact) % MOD;
  long long up = pow(down, MOD - 2);
  return (n_fact * up) % MOD;
}
long long answer = 0;
long long total_inv = 0;
long long n;
vector<pair<long long, long long>> LEVELS(33, make_pair(0, 0));
inline void print(const vector<int>& bits) {
  long long inv = 0, ones = 0;
  cerr << "BITS: ";
  for (int i = 0; i < bits.size(); i++) {
    cerr << bits[i];
  }
  cerr << endl;
}
inline long long inv_count(const vector<int>& bits) {
  long long inv = 0, ones = 0;
  for (int i = 0; i < bits.size(); i++) {
    if (bits[i] == 1)
      ones++;
    else
      inv += ones;
  }
  return inv;
}
inline long long inv_count2(const vector<int>& bits) {
  long long inv = 0, zeros = 0;
  for (int i = 0; i < bits.size(); i++) {
    if (bits[i] == 0)
      zeros++;
    else
      inv += zeros;
  }
  return inv;
}
void dfs(vector<long long> vec, long long level) {
  vector<int> bits;
  vector<long long> ones, zeros;
  long long BIT = (1 << level);
  for (int i = 0; i < vec.size(); i++) {
    if (vec[i] & BIT) {
      bits.push_back(1);
      ones.push_back(vec[i]);
    } else {
      bits.push_back(0);
      zeros.push_back(vec[i]);
    }
  }
  long long q = inv_count(bits);
  long long q1 = inv_count2(bits);
  LEVELS[level].first += q;
  LEVELS[level].second += q1;
  if (level > 0) {
    if (!ones.empty()) dfs(ones, level - 1);
    if (!zeros.empty()) dfs(zeros, level - 1);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  vector<long long> vec(n);
  for (int i = 0; i < n; i++) {
    cin >> vec[i];
  }
  dfs(vec, 31);
  for (long long level = 0; level <= 31; level++) {
    long long BIT = (1 << level);
    if (LEVELS[level].second < LEVELS[level].first) {
      answer |= BIT;
      total_inv += LEVELS[level].second;
    } else {
      total_inv += LEVELS[level].first;
    }
  }
  cout << total_inv << " " << answer << endl;
  return 0;
}

```