#include <bits/stdc++.h>
using namespace std;
bool compare(const pair<int long long, int long long>& a,
             const pair<int long long, int long long>& b) {
  return a.second <= b.second;
}
int long long inline power(int long long a, int long long b, int long long p) {
  a %= p;
  int long long ans = 1;
  while (b > 0) {
    if (b & 1) ans = (ans * a) % p;
    a = (a * a) % p;
    b >>= 1;
  }
  return ans;
}
int long long inv(int long long n, int long long p) {
  return power(n, p - 2, p);
}
bool inline isPrime(int long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (int long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
int long long gcd(int long long a, int long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int long long dp0[50], dp1[50];
struct trie {
  struct node {
    node* nxt[2];
    int long long cnt;
    node() {
      cnt = 0;
      nxt[0] = nxt[1] = NULL;
    }
  };
  node* root;
  trie() { root = new node(); }
  void insert(int long long x) {
    node* cur = root;
    for (int long long i = 31; i >= 0; i--) {
      int long long b = (x >> i) & 1;
      if (b && cur->nxt[0])
        dp1[i] += cur->nxt[0]->cnt;
      else if (!b && cur->nxt[1])
        dp0[i] += cur->nxt[1]->cnt;
      if (!cur->nxt[b]) {
        cur->nxt[b] = new node();
      }
      cur = cur->nxt[b];
      cur->cnt++;
    }
  }
  void remove(int long long x) {
    node* cur = root;
    for (int long long i = 31; i >= 0; i--) {
      int long long b = (x >> i) & 1;
      cur = cur->nxt[b];
      cur->cnt--;
    }
  }
  int long long maxxor(int long long x) {
    node* cur = root;
    int long long ans = 0;
    for (int long long i = 31; i >= 0; i--) {
      int long long b = (x >> i) & 1;
      if (cur->nxt[!b] && cur->nxt[!b]->cnt > 0) {
        ans += (1LL << i);
        cur = cur->nxt[!b];
      } else
        cur = cur->nxt[b];
    }
    return ans;
  }
};
void solve() {
  int long long n, m, i, j, k, l, r;
  cin >> n;
  trie T;
  memset(dp0, 0, sizeof(dp0));
  memset(dp1, 0, sizeof(dp1));
  for (i = 0; i < n; i++) {
    cin >> l;
    T.insert(l);
  }
  int long long inv = 0, val = 0;
  for (i = 31; i >= 0; i--) {
    inv += min(dp0[i], dp1[i]);
    if (dp1[i] < dp0[i]) val += 1LL << i;
  }
  cout << inv << " " << val << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int long long t = 1;
  int long long w = 1;
  while (t--) {
    solve();
    w++;
  }
}

### Prompt

In CPP, your task is to solve 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;
bool compare(const pair<int long long, int long long>& a,
             const pair<int long long, int long long>& b) {
  return a.second <= b.second;
}
int long long inline power(int long long a, int long long b, int long long p) {
  a %= p;
  int long long ans = 1;
  while (b > 0) {
    if (b & 1) ans = (ans * a) % p;
    a = (a * a) % p;
    b >>= 1;
  }
  return ans;
}
int long long inv(int long long n, int long long p) {
  return power(n, p - 2, p);
}
bool inline isPrime(int long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (int long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
int long long gcd(int long long a, int long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int long long dp0[50], dp1[50];
struct trie {
  struct node {
    node* nxt[2];
    int long long cnt;
    node() {
      cnt = 0;
      nxt[0] = nxt[1] = NULL;
    }
  };
  node* root;
  trie() { root = new node(); }
  void insert(int long long x) {
    node* cur = root;
    for (int long long i = 31; i >= 0; i--) {
      int long long b = (x >> i) & 1;
      if (b && cur->nxt[0])
        dp1[i] += cur->nxt[0]->cnt;
      else if (!b && cur->nxt[1])
        dp0[i] += cur->nxt[1]->cnt;
      if (!cur->nxt[b]) {
        cur->nxt[b] = new node();
      }
      cur = cur->nxt[b];
      cur->cnt++;
    }
  }
  void remove(int long long x) {
    node* cur = root;
    for (int long long i = 31; i >= 0; i--) {
      int long long b = (x >> i) & 1;
      cur = cur->nxt[b];
      cur->cnt--;
    }
  }
  int long long maxxor(int long long x) {
    node* cur = root;
    int long long ans = 0;
    for (int long long i = 31; i >= 0; i--) {
      int long long b = (x >> i) & 1;
      if (cur->nxt[!b] && cur->nxt[!b]->cnt > 0) {
        ans += (1LL << i);
        cur = cur->nxt[!b];
      } else
        cur = cur->nxt[b];
    }
    return ans;
  }
};
void solve() {
  int long long n, m, i, j, k, l, r;
  cin >> n;
  trie T;
  memset(dp0, 0, sizeof(dp0));
  memset(dp1, 0, sizeof(dp1));
  for (i = 0; i < n; i++) {
    cin >> l;
    T.insert(l);
  }
  int long long inv = 0, val = 0;
  for (i = 31; i >= 0; i--) {
    inv += min(dp0[i], dp1[i]);
    if (dp1[i] < dp0[i]) val += 1LL << i;
  }
  cout << inv << " " << val << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int long long t = 1;
  int long long w = 1;
  while (t--) {
    solve();
    w++;
  }
}

```