#include <bits/stdc++.h>
using namespace std;
vector<long long> p(300000);
vector<long long> sz(300000, 1);
long long root(long long v) {
  if (v != p[v]) p[v] = root(p[v]);
  return p[v];
}
bool unite(long long a, long long b) {
  a = root(a);
  b = root(b);
  if (a == b) return 0;
  if (sz[a] < sz[b]) {
    p[a] = b;
    sz[b] += sz[a];
  } else {
    p[b] = a;
    sz[a] += sz[b];
  }
  return 1;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  vector<long long> a(n), zeros(n), ones(n);
  vector<pair<long long, long long>> s;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    s.emplace_back(a[i], i);
    p[i] = i;
  }
  sort(s.begin(), s.end());
  for (long long i = 1; i < n; i++) {
    if (s[i].first == s[i - 1].first) {
      unite(s[i].second, s[i - 1].second);
    }
  }
  long long ans = 0;
  long long x = 0;
  for (long long bit = 0; bit <= 30; bit++) {
    s.clear();
    long long s0 = 0;
    long long s1 = 0;
    for (long long i = 0; i < n; i++) {
      long long pref = a[i] >> (bit + 1ll);
      s.emplace_back(pref, i);
    }
    sort(s.begin(), s.end());
    for (long long i = 1; i < n; i++) {
      if (s[i].first == s[i - 1].first) {
        unite(s[i].second, s[i - 1].second);
      }
    }
    for (long long i = 0; i < n; i++) {
      long long rt = root(i);
      if (a[i] & (1ll << bit)) {
        s1 += zeros[rt];
        ones[rt]++;
      } else {
        s0 += ones[rt];
        zeros[rt]++;
      }
    }
    for (long long i = 0; i < n; i++) {
      ones[i] = 0;
      zeros[i] = 0;
    }
    if (s0 <= s1) {
      ans += s0;
    } else {
      ans += s1;
      x += (1ll << bit);
    }
  }
  cout << ans << ' ' << x;
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
vector<long long> p(300000);
vector<long long> sz(300000, 1);
long long root(long long v) {
  if (v != p[v]) p[v] = root(p[v]);
  return p[v];
}
bool unite(long long a, long long b) {
  a = root(a);
  b = root(b);
  if (a == b) return 0;
  if (sz[a] < sz[b]) {
    p[a] = b;
    sz[b] += sz[a];
  } else {
    p[b] = a;
    sz[a] += sz[b];
  }
  return 1;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  vector<long long> a(n), zeros(n), ones(n);
  vector<pair<long long, long long>> s;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    s.emplace_back(a[i], i);
    p[i] = i;
  }
  sort(s.begin(), s.end());
  for (long long i = 1; i < n; i++) {
    if (s[i].first == s[i - 1].first) {
      unite(s[i].second, s[i - 1].second);
    }
  }
  long long ans = 0;
  long long x = 0;
  for (long long bit = 0; bit <= 30; bit++) {
    s.clear();
    long long s0 = 0;
    long long s1 = 0;
    for (long long i = 0; i < n; i++) {
      long long pref = a[i] >> (bit + 1ll);
      s.emplace_back(pref, i);
    }
    sort(s.begin(), s.end());
    for (long long i = 1; i < n; i++) {
      if (s[i].first == s[i - 1].first) {
        unite(s[i].second, s[i - 1].second);
      }
    }
    for (long long i = 0; i < n; i++) {
      long long rt = root(i);
      if (a[i] & (1ll << bit)) {
        s1 += zeros[rt];
        ones[rt]++;
      } else {
        s0 += ones[rt];
        zeros[rt]++;
      }
    }
    for (long long i = 0; i < n; i++) {
      ones[i] = 0;
      zeros[i] = 0;
    }
    if (s0 <= s1) {
      ans += s0;
    } else {
      ans += s1;
      x += (1ll << bit);
    }
  }
  cout << ans << ' ' << x;
  return 0;
}

```