#include <bits/stdc++.h>
using namespace std;
long long byte_size = 29;
vector<vector<long long>> trie;
vector<string> st;
vector<vector<long long>> sum(byte_size + 2, vector<long long>(2, 0));
long long depth_check = 0;
long long bin_to_int(string s) {
  long long k = 1;
  long long ans = 0;
  for (long long i = s.length() - 1; i >= 0; i--) {
    ans += (s[i] - '0') * k;
    k *= 2;
  }
  return ans;
}
void add(long long x) {
  long long v = 0;
  for (long long i = byte_size; i >= 0; i--) {
    long long bit = (x & (1 << i)) != 0;
    if (trie[v][bit] == -1) {
      trie[v][bit] = trie.size();
      trie.emplace_back(vector<long long>(2, -1));
      st.emplace_back();
    }
    st[v] += to_string(bit);
    v = trie[v][bit];
  }
}
string inverse(string s) {
  for (long long i = 0; i < s.length(); i++) {
    if (s[i] == '0')
      s[i] = '1';
    else
      s[i] = '0';
  }
  return s;
}
long long solve(string s) {
  long long cnt = 0;
  long long ans = 0;
  for (long long i = (long long)s.length() - 1; i >= 0; i--) {
    if (s[i] == '0')
      cnt++;
    else {
      ans += cnt;
    }
  }
  return ans;
}
void dfs(long long v, long long depth = 0) {
  if (v == -1 || (trie[v][0] == -1 && trie[v][1] == -1)) return;
  depth_check = max(depth_check, depth);
  string h = st[v];
  sum[depth][0] += solve(h);
  sum[depth][1] += solve(inverse(h));
  dfs(trie[v][0], depth + 1), dfs(trie[v][1], depth + 1);
}
signed main() {
  long long n;
  cin >> n;
  vector<long long> a(n, 0);
  for (long long i = 0; i < n; i++) cin >> a[i];
  trie.assign(1, vector<long long>(2, -1));
  for (long long i = 0; i < n; i++) {
    add(a[i]);
  }
  dfs(0);
  string s;
  long long ans = 0;
  for (long long i = 0; i < sum.size(); i++) {
    if (i > depth_check) break;
    if (sum[i][0] <= sum[i][1]) {
      s += '0';
    } else
      s += '1';
    ans += sum[i][sum[i][0] > sum[i][1]];
  }
  cout << ans << " " << bin_to_int(s) << "\n";
}

### 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;
long long byte_size = 29;
vector<vector<long long>> trie;
vector<string> st;
vector<vector<long long>> sum(byte_size + 2, vector<long long>(2, 0));
long long depth_check = 0;
long long bin_to_int(string s) {
  long long k = 1;
  long long ans = 0;
  for (long long i = s.length() - 1; i >= 0; i--) {
    ans += (s[i] - '0') * k;
    k *= 2;
  }
  return ans;
}
void add(long long x) {
  long long v = 0;
  for (long long i = byte_size; i >= 0; i--) {
    long long bit = (x & (1 << i)) != 0;
    if (trie[v][bit] == -1) {
      trie[v][bit] = trie.size();
      trie.emplace_back(vector<long long>(2, -1));
      st.emplace_back();
    }
    st[v] += to_string(bit);
    v = trie[v][bit];
  }
}
string inverse(string s) {
  for (long long i = 0; i < s.length(); i++) {
    if (s[i] == '0')
      s[i] = '1';
    else
      s[i] = '0';
  }
  return s;
}
long long solve(string s) {
  long long cnt = 0;
  long long ans = 0;
  for (long long i = (long long)s.length() - 1; i >= 0; i--) {
    if (s[i] == '0')
      cnt++;
    else {
      ans += cnt;
    }
  }
  return ans;
}
void dfs(long long v, long long depth = 0) {
  if (v == -1 || (trie[v][0] == -1 && trie[v][1] == -1)) return;
  depth_check = max(depth_check, depth);
  string h = st[v];
  sum[depth][0] += solve(h);
  sum[depth][1] += solve(inverse(h));
  dfs(trie[v][0], depth + 1), dfs(trie[v][1], depth + 1);
}
signed main() {
  long long n;
  cin >> n;
  vector<long long> a(n, 0);
  for (long long i = 0; i < n; i++) cin >> a[i];
  trie.assign(1, vector<long long>(2, -1));
  for (long long i = 0; i < n; i++) {
    add(a[i]);
  }
  dfs(0);
  string s;
  long long ans = 0;
  for (long long i = 0; i < sum.size(); i++) {
    if (i > depth_check) break;
    if (sum[i][0] <= sum[i][1]) {
      s += '0';
    } else
      s += '1';
    ans += sum[i][sum[i][0] > sum[i][1]];
  }
  cout << ans << " " << bin_to_int(s) << "\n";
}

```