#include <bits/stdc++.h>
using namespace std;
const int64_t INF = 2e9;
struct SEG {
  int64_t sz;
  vector<int64_t> a;
  SEG(int64_t n) {
    for (sz = 1; sz < n; sz *= 2)
      ;
    a.resize(2 * sz);
  }
  void update(int64_t i) {
    a[i += sz]++;
    for (i /= 2; i; i /= 2) a[i] = a[2 * i] + a[2 * i + 1];
  }
  int64_t sum(int64_t l, int64_t r) {
    int64_t res = 0;
    for (l += sz, r += sz; l <= r; l /= 2, r /= 2) {
      if (l % 2) res += a[l++];
      if (r % 2 == 0) res += a[r--];
    }
    return res;
  }
};
int64_t n;
int64_t getv(vector<int64_t>& a) {
  map<int64_t, int64_t> mp;
  for (int64_t i = 0; i < n; i++) mp[a[i]];
  int64_t cnt = 0;
  for (auto& first : mp) first.second = cnt++;
  SEG seg(cnt + 1);
  int64_t ans = 0;
  for (int64_t i = 0; i < n; i++) {
    int64_t first = mp[a[i]];
    ans += seg.sum(first + 1, cnt);
    seg.update(first);
  }
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  vector<int64_t> a(n);
  for (int64_t i = 0; i < n; i++) cin >> a[i];
  int64_t first = 0;
  for (int64_t k = 32 - 1; k >= 0; k--) {
    vector<pair<int64_t, int64_t> > top(n);
    for (int64_t i = 0; i < n; i++) {
      top[i] = {a[i] >> (k + 1), i};
    }
    sort(top.begin(), top.end());
    int64_t v1 = 0, v2 = 0, last = -1;
    vector<int64_t> cur;
    for (auto p : top) {
      if (p.first != last) {
        int64_t z = 0, o = 0;
        for (auto i : cur) {
          int64_t v = a[i] & (1 << k);
          if (v)
            o++, v2 += z;
          else
            z++, v1 += o;
        }
        cur.clear();
      }
      cur.push_back(p.second);
      last = p.first;
    }
    int64_t z = 0, o = 0;
    for (auto i : cur) {
      int64_t v = a[i] & (1 << k);
      if (v)
        o++, v2 += z;
      else
        z++, v1 += o;
    }
    cur.clear();
    if (v1 > v2) {
      first ^= (1 << k);
      for (int64_t i = 0; i < n; i++) a[i] ^= (1 << k);
    }
  }
  cout << getv(a) << " " << first << endl;
  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>
using namespace std;
const int64_t INF = 2e9;
struct SEG {
  int64_t sz;
  vector<int64_t> a;
  SEG(int64_t n) {
    for (sz = 1; sz < n; sz *= 2)
      ;
    a.resize(2 * sz);
  }
  void update(int64_t i) {
    a[i += sz]++;
    for (i /= 2; i; i /= 2) a[i] = a[2 * i] + a[2 * i + 1];
  }
  int64_t sum(int64_t l, int64_t r) {
    int64_t res = 0;
    for (l += sz, r += sz; l <= r; l /= 2, r /= 2) {
      if (l % 2) res += a[l++];
      if (r % 2 == 0) res += a[r--];
    }
    return res;
  }
};
int64_t n;
int64_t getv(vector<int64_t>& a) {
  map<int64_t, int64_t> mp;
  for (int64_t i = 0; i < n; i++) mp[a[i]];
  int64_t cnt = 0;
  for (auto& first : mp) first.second = cnt++;
  SEG seg(cnt + 1);
  int64_t ans = 0;
  for (int64_t i = 0; i < n; i++) {
    int64_t first = mp[a[i]];
    ans += seg.sum(first + 1, cnt);
    seg.update(first);
  }
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  vector<int64_t> a(n);
  for (int64_t i = 0; i < n; i++) cin >> a[i];
  int64_t first = 0;
  for (int64_t k = 32 - 1; k >= 0; k--) {
    vector<pair<int64_t, int64_t> > top(n);
    for (int64_t i = 0; i < n; i++) {
      top[i] = {a[i] >> (k + 1), i};
    }
    sort(top.begin(), top.end());
    int64_t v1 = 0, v2 = 0, last = -1;
    vector<int64_t> cur;
    for (auto p : top) {
      if (p.first != last) {
        int64_t z = 0, o = 0;
        for (auto i : cur) {
          int64_t v = a[i] & (1 << k);
          if (v)
            o++, v2 += z;
          else
            z++, v1 += o;
        }
        cur.clear();
      }
      cur.push_back(p.second);
      last = p.first;
    }
    int64_t z = 0, o = 0;
    for (auto i : cur) {
      int64_t v = a[i] & (1 << k);
      if (v)
        o++, v2 += z;
      else
        z++, v1 += o;
    }
    cur.clear();
    if (v1 > v2) {
      first ^= (1 << k);
      for (int64_t i = 0; i < n; i++) a[i] ^= (1 << k);
    }
  }
  cout << getv(a) << " " << first << endl;
  return 0;
}

```