#include <bits/stdc++.h>
using namespace std;
struct BIT {
  int n;
  vector<long long> bit;
  void init(int _n) {
    n = _n;
    bit.assign(n + 1, 0);
  }
  long long q(int x) {
    long long ret = 0;
    while (x > 0) ret += bit[x], x -= x & (-x);
    return ret;
  }
  void m(int x, long long d) {
    while (x <= n) bit[x] += d, x += x & (-x);
  }
};
void marmot0814() {
  long long n;
  cin >> n;
  vector<int> arr(n);
  for (auto &v : arr) cin >> v;
  long long ans = 0;
  queue<vector<int>> q;
  vector<int> base;
  for (int i = 0; i < n; i++) base.push_back(i);
  q.push(base);
  for (int i = 29; i >= 0; i--) {
    int t = q.size();
    long long sum = 0, tot = 0;
    while (t--) {
      auto idx = q.front();
      q.pop();
      vector<int> res;
      for (auto &v : idx) res.push_back((arr[v] >> i) & 1);
      long long one = 0, inv = 0;
      for (int j = 0; j < (int)res.size(); j++) {
        if (res[j] == 0) inv += one;
        one += res[j];
      }
      sum += inv;
      tot += one * ((long long)res.size() - one);
      vector<int> Z, O;
      for (auto &v : idx) {
        if ((arr[v] >> i) & 1)
          O.push_back(v);
        else
          Z.push_back(v);
      }
      if (O.size() > 1) q.push(O);
      if (Z.size() > 1) q.push(Z);
    }
    long long bef = sum;
    long long aft = tot - sum;
    if (aft < bef) ans ^= (1 << i);
  }
  map<int, int> mp;
  for (auto &v : arr) {
    v ^= ans;
    mp[v];
  }
  int st = 0;
  for (auto &p : mp) p.second = ++st;
  for (auto &v : arr) v = mp[v];
  BIT bit;
  bit.init(mp.size());
  long long res = 0;
  for (int i = 0; i < n; i++) {
    res += bit.q(mp.size()) - bit.q(arr[i]);
    bit.m(arr[i], 1);
  }
  cout << res << ' ' << ans << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t = 1, kase = 0;
  while (t--) {
    marmot0814();
  }
}

### Prompt

Please create a solution in CPP 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>
using namespace std;
struct BIT {
  int n;
  vector<long long> bit;
  void init(int _n) {
    n = _n;
    bit.assign(n + 1, 0);
  }
  long long q(int x) {
    long long ret = 0;
    while (x > 0) ret += bit[x], x -= x & (-x);
    return ret;
  }
  void m(int x, long long d) {
    while (x <= n) bit[x] += d, x += x & (-x);
  }
};
void marmot0814() {
  long long n;
  cin >> n;
  vector<int> arr(n);
  for (auto &v : arr) cin >> v;
  long long ans = 0;
  queue<vector<int>> q;
  vector<int> base;
  for (int i = 0; i < n; i++) base.push_back(i);
  q.push(base);
  for (int i = 29; i >= 0; i--) {
    int t = q.size();
    long long sum = 0, tot = 0;
    while (t--) {
      auto idx = q.front();
      q.pop();
      vector<int> res;
      for (auto &v : idx) res.push_back((arr[v] >> i) & 1);
      long long one = 0, inv = 0;
      for (int j = 0; j < (int)res.size(); j++) {
        if (res[j] == 0) inv += one;
        one += res[j];
      }
      sum += inv;
      tot += one * ((long long)res.size() - one);
      vector<int> Z, O;
      for (auto &v : idx) {
        if ((arr[v] >> i) & 1)
          O.push_back(v);
        else
          Z.push_back(v);
      }
      if (O.size() > 1) q.push(O);
      if (Z.size() > 1) q.push(Z);
    }
    long long bef = sum;
    long long aft = tot - sum;
    if (aft < bef) ans ^= (1 << i);
  }
  map<int, int> mp;
  for (auto &v : arr) {
    v ^= ans;
    mp[v];
  }
  int st = 0;
  for (auto &p : mp) p.second = ++st;
  for (auto &v : arr) v = mp[v];
  BIT bit;
  bit.init(mp.size());
  long long res = 0;
  for (int i = 0; i < n; i++) {
    res += bit.q(mp.size()) - bit.q(arr[i]);
    bit.m(arr[i], 1);
  }
  cout << res << ' ' << ans << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t = 1, kase = 0;
  while (t--) {
    marmot0814();
  }
}

```