#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
void __print(int x) { cerr << x; }
void __print(long long int x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char* x) { cerr << '\"' << x << '\"'; }
void __print(const string& x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V>& x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T& x) {
  int _ = 0;
  cerr << '{';
  for (auto& i : x) cerr << (_++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
long long int INF = 2e17;
long long int MOD = 1e9 + 7;
template <typename T>
struct INVCOUNT {
  long long int inv = 0;
  pair<long long int, long long int> init(vector<T>& a) {
    int n = a.size();
    vector<int> pre(1);
    for (auto it : a) {
      pre.push_back(pre.back() + it);
    }
    long long int norm = 0;
    long long int flip = 0;
    for (int i = 0; i < n; ++i) {
      if (a[i]) {
        int zeros = i - pre[i];
        flip += zeros;
      } else {
        int ones = pre[i];
        norm += ones;
      }
    }
    return {norm, flip};
  }
};
void solve_tc(int test_case) {
  int n;
  cin >> n;
  vector<int> a(n);
  vector<int> p(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    p[i] = i;
  }
  long long int total = 0;
  vector<pair<int, vector<int>>> cur;
  cur.emplace_back(31, p);
  int num = 0;
  for (int i = 31; i >= 0; --i) {
    vector<pair<int, vector<int>>> now;
    long long int inv = 0;
    long long int invFlip = 0;
    for (auto x : cur) {
      vector<int> needto;
      for (auto it : x.second) {
        needto.push_back((a[it] & (1 << x.first)) > 0);
      }
      INVCOUNT<int> invCount;
      auto [invHere, invHereFlip] = invCount.init(needto);
      inv += invHere;
      invFlip += invHereFlip;
      if (i) {
        int ndep = i - 1;
        pair<int, vector<int>> l, r;
        l.first = r.first = ndep;
        for (auto it : x.second) {
          if (a[it] & (1 << i))
            l.second.push_back(it);
          else
            r.second.push_back(it);
        }
        if (l.second.size()) {
          now.push_back(l);
        }
        if (r.second.size()) {
          now.push_back(r);
        }
      }
    }
    if (inv <= invFlip) {
      total += inv;
    } else {
      total += invFlip;
      num += (1 << i);
    }
    swap(cur, now);
  }
  cout << total << ' ' << num << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  auto start_exec = chrono::high_resolution_clock::now();
  int tt = 1;
  for (int tc = 1; tc <= tt; ++tc) {
    solve_tc(tc);
  }
  auto stop_exec = chrono::high_resolution_clock::now();
  auto duration_exec =
      chrono::duration_cast<chrono::milliseconds>(stop_exec - start_exec);
  return 0;
}

### Prompt

Your challenge is to write 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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
void __print(int x) { cerr << x; }
void __print(long long int x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char* x) { cerr << '\"' << x << '\"'; }
void __print(const string& x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V>& x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T& x) {
  int _ = 0;
  cerr << '{';
  for (auto& i : x) cerr << (_++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
long long int INF = 2e17;
long long int MOD = 1e9 + 7;
template <typename T>
struct INVCOUNT {
  long long int inv = 0;
  pair<long long int, long long int> init(vector<T>& a) {
    int n = a.size();
    vector<int> pre(1);
    for (auto it : a) {
      pre.push_back(pre.back() + it);
    }
    long long int norm = 0;
    long long int flip = 0;
    for (int i = 0; i < n; ++i) {
      if (a[i]) {
        int zeros = i - pre[i];
        flip += zeros;
      } else {
        int ones = pre[i];
        norm += ones;
      }
    }
    return {norm, flip};
  }
};
void solve_tc(int test_case) {
  int n;
  cin >> n;
  vector<int> a(n);
  vector<int> p(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    p[i] = i;
  }
  long long int total = 0;
  vector<pair<int, vector<int>>> cur;
  cur.emplace_back(31, p);
  int num = 0;
  for (int i = 31; i >= 0; --i) {
    vector<pair<int, vector<int>>> now;
    long long int inv = 0;
    long long int invFlip = 0;
    for (auto x : cur) {
      vector<int> needto;
      for (auto it : x.second) {
        needto.push_back((a[it] & (1 << x.first)) > 0);
      }
      INVCOUNT<int> invCount;
      auto [invHere, invHereFlip] = invCount.init(needto);
      inv += invHere;
      invFlip += invHereFlip;
      if (i) {
        int ndep = i - 1;
        pair<int, vector<int>> l, r;
        l.first = r.first = ndep;
        for (auto it : x.second) {
          if (a[it] & (1 << i))
            l.second.push_back(it);
          else
            r.second.push_back(it);
        }
        if (l.second.size()) {
          now.push_back(l);
        }
        if (r.second.size()) {
          now.push_back(r);
        }
      }
    }
    if (inv <= invFlip) {
      total += inv;
    } else {
      total += invFlip;
      num += (1 << i);
    }
    swap(cur, now);
  }
  cout << total << ' ' << num << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  auto start_exec = chrono::high_resolution_clock::now();
  int tt = 1;
  for (int tc = 1; tc <= tt; ++tc) {
    solve_tc(tc);
  }
  auto stop_exec = chrono::high_resolution_clock::now();
  auto duration_exec =
      chrono::duration_cast<chrono::milliseconds>(stop_exec - start_exec);
  return 0;
}

```