#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
long long expo_pow(long long x, long long y) {
  if (y == 0) return 1;
  y = y % (mod - 1);
  x %= mod;
  if (y == 0) y = mod - 1;
  long long res = 1;
  while (y) {
    if (y & 1) res = (res * x) % mod;
    x = (x * x) % mod;
    y >>= 1;
  }
  return res;
}
long long add() { return 0; }
template <typename T, typename... Types>
T add(T var1, Types... var2) {
  return (((((long long)(var1)) % mod + (long long)(add(var2...))) % mod) +
          mod) %
         mod;
}
long long mul() { return 1; }
template <typename T, typename... Types>
T mul(T var1, Types... var2) {
  return (((long long)(var1)) % mod * (long long)(mul(var2...))) % mod;
}
const long long K = 2;
struct Vertex {
  long long next[K];
  bool leaf = false;
  vector<long long> index;
  long long inv = 0;
  Vertex() { fill(begin(next), end(next), -1); }
};
vector<Vertex> trie(1);
void add_string(string const& second, long long idx) {
  long long v = 0;
  vector<long long> path;
  path.push_back(0);
  for (char ch : second) {
    long long c = ch - '0';
    if (trie[v].next[c] == -1) {
      trie[v].next[c] = trie.size();
      trie.emplace_back();
    }
    v = trie[v].next[c];
    path.push_back(v);
  }
  trie[v].leaf = true;
  trie[v].index.push_back(idx);
  for (long long i = path.size() - 2; i >= 0; i--) {
    trie[path[i]].index.push_back(idx);
  }
}
string get(long long tmp) {
  string ans;
  while (tmp) {
    ans.push_back('0' + tmp % 2);
    tmp /= 2;
  }
  while (ans.size() != 31) ans.push_back('0');
  reverse(ans.begin(), ans.end());
  return ans;
}
long long sum[31][2];
long long get_inv(vector<long long> v1, vector<long long> v2) {
  long long i = 0;
  long long j = 0;
  long long ans = 0;
  while (i < v1.size() and j < v2.size()) {
    if (v1[i] > v2[j]) {
      ans += v1.size() - i;
      j++;
    } else
      i++;
  }
  return ans;
}
void dfs(long long cur, long long height = 30, long long prev_sum = 0) {
  vector<long long> v1, v2;
  if (trie[cur].next[0] != -1) {
    v1 = trie[trie[cur].next[0]].index;
  }
  if (trie[cur].next[1] != -1) {
    v2 = trie[trie[cur].next[1]].index;
  }
  if (!v1.empty() and !v2.empty()) {
    long long inv1 = get_inv(v1, v2);
    long long inv2 = get_inv(v2, v1);
    sum[height][0] += inv1;
    sum[height][1] += inv2;
    dfs(trie[cur].next[1], height - 1);
    dfs(trie[cur].next[0], height - 1);
  } else if (!v1.empty()) {
    dfs(trie[cur].next[0], height - 1);
  } else if (!v2.empty()) {
    dfs(trie[cur].next[1], height - 1);
  }
}
void solve() {
  long long n;
  cin >> n;
  for (long long i = 0; i < n; ++i) {
    long long tmp;
    cin >> tmp;
    string bin = get(tmp);
    add_string(bin, i);
  }
  dfs(0);
  long long ans = 0;
  long long inv = 0;
  for (long long i = 30; i >= 0; --i) {
    if (sum[i][0] <= sum[i][1]) {
      inv += sum[i][0];
    } else {
      ans |= (1 << i);
      inv += sum[i][1];
    }
  }
  cout << inv << " " << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long test = 1;
  long long i = 1;
  while (test--) {
    solve();
  }
}

### 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;
const long long mod = 1e9 + 7;
long long expo_pow(long long x, long long y) {
  if (y == 0) return 1;
  y = y % (mod - 1);
  x %= mod;
  if (y == 0) y = mod - 1;
  long long res = 1;
  while (y) {
    if (y & 1) res = (res * x) % mod;
    x = (x * x) % mod;
    y >>= 1;
  }
  return res;
}
long long add() { return 0; }
template <typename T, typename... Types>
T add(T var1, Types... var2) {
  return (((((long long)(var1)) % mod + (long long)(add(var2...))) % mod) +
          mod) %
         mod;
}
long long mul() { return 1; }
template <typename T, typename... Types>
T mul(T var1, Types... var2) {
  return (((long long)(var1)) % mod * (long long)(mul(var2...))) % mod;
}
const long long K = 2;
struct Vertex {
  long long next[K];
  bool leaf = false;
  vector<long long> index;
  long long inv = 0;
  Vertex() { fill(begin(next), end(next), -1); }
};
vector<Vertex> trie(1);
void add_string(string const& second, long long idx) {
  long long v = 0;
  vector<long long> path;
  path.push_back(0);
  for (char ch : second) {
    long long c = ch - '0';
    if (trie[v].next[c] == -1) {
      trie[v].next[c] = trie.size();
      trie.emplace_back();
    }
    v = trie[v].next[c];
    path.push_back(v);
  }
  trie[v].leaf = true;
  trie[v].index.push_back(idx);
  for (long long i = path.size() - 2; i >= 0; i--) {
    trie[path[i]].index.push_back(idx);
  }
}
string get(long long tmp) {
  string ans;
  while (tmp) {
    ans.push_back('0' + tmp % 2);
    tmp /= 2;
  }
  while (ans.size() != 31) ans.push_back('0');
  reverse(ans.begin(), ans.end());
  return ans;
}
long long sum[31][2];
long long get_inv(vector<long long> v1, vector<long long> v2) {
  long long i = 0;
  long long j = 0;
  long long ans = 0;
  while (i < v1.size() and j < v2.size()) {
    if (v1[i] > v2[j]) {
      ans += v1.size() - i;
      j++;
    } else
      i++;
  }
  return ans;
}
void dfs(long long cur, long long height = 30, long long prev_sum = 0) {
  vector<long long> v1, v2;
  if (trie[cur].next[0] != -1) {
    v1 = trie[trie[cur].next[0]].index;
  }
  if (trie[cur].next[1] != -1) {
    v2 = trie[trie[cur].next[1]].index;
  }
  if (!v1.empty() and !v2.empty()) {
    long long inv1 = get_inv(v1, v2);
    long long inv2 = get_inv(v2, v1);
    sum[height][0] += inv1;
    sum[height][1] += inv2;
    dfs(trie[cur].next[1], height - 1);
    dfs(trie[cur].next[0], height - 1);
  } else if (!v1.empty()) {
    dfs(trie[cur].next[0], height - 1);
  } else if (!v2.empty()) {
    dfs(trie[cur].next[1], height - 1);
  }
}
void solve() {
  long long n;
  cin >> n;
  for (long long i = 0; i < n; ++i) {
    long long tmp;
    cin >> tmp;
    string bin = get(tmp);
    add_string(bin, i);
  }
  dfs(0);
  long long ans = 0;
  long long inv = 0;
  for (long long i = 30; i >= 0; --i) {
    if (sum[i][0] <= sum[i][1]) {
      inv += sum[i][0];
    } else {
      ans |= (1 << i);
      inv += sum[i][1];
    }
  }
  cout << inv << " " << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long test = 1;
  long long i = 1;
  while (test--) {
    solve();
  }
}

```