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.
#include <bits/stdc++.h>
using namespace std;
template <class T1, class T2>
void pr(const pair<T1, T2>& x);
template <class T, size_t SZ>
void pr(const array<T, SZ>& x);
template <class T>
void pr(const vector<T>& x);
template <class T>
void pr(const set<T>& x);
template <class T1, class T2>
void pr(const map<T1, T2>& x);
template <class T>
void pr(const T& x) {
  cout << x;
}
template <class Arg, class... Args>
void pr(const Arg& first, const Args&... rest) {
  pr(first);
  pr(rest...);
}
template <class T1, class T2>
void pr(const pair<T1, T2>& x) {
  pr("{", x.first, ", ", x.second, "}");
}
template <class T>
void prContain(const T& x) {
  pr("{");
  bool fst = 1;
  for (auto& a : x) pr(!fst ? ", " : "", a), fst = 0;
  pr("}");
}
template <class T, size_t SZ>
void pr(const array<T, SZ>& x) {
  prContain(x);
}
template <class T>
void pr(const vector<T>& x) {
  prContain(x);
}
template <class T>
void pr(const set<T>& x) {
  prContain(x);
}
template <class T1, class T2>
void pr(const map<T1, T2>& x) {
  prContain(x);
}
void ps() { pr("\n"); }
template <class Arg, class... Args>
void ps(const Arg& first, const Args&... rest) {
  pr(first, " ");
  ps(rest...);
}
template <typename T>
int remin(T& a, const T& b) {
  if (b < a) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
int remax(T& a, const T& b) {
  if (b > a) {
    a = b;
    return true;
  }
  return false;
}
int n;
const int BIN = 2;
const int DEPTH = 29;
const int NN = 6000105;
vector<int> ind[NN][2];
long long dp[30][2];
vector<vector<int> > trie;
void dfs(int u = 0, int depth = DEPTH) {
  int l = trie[u][0], r = trie[u][1];
  if (l != -1) dfs(l, depth - 1);
  if (r != -1) dfs(r, depth - 1);
  if (l == -1 || r == -1) return;
  vector<int> gl = ind[u][0], gr = ind[u][1];
  int nl = gl.size(), nr = gr.size();
  long long cnt = 0;
  for (int k : gl) {
    cnt += lower_bound(gr.begin(), gr.end(), k) - gr.begin();
  }
  dp[depth][0] += cnt;
  dp[depth][1] += 1LL * nl * nr - cnt;
}
void solve() {
  scanf("%d", &n);
  vector<int> v(n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &v[i]);
  }
  trie.assign(1, vector<int>(BIN, -1));
  int pt_mx = 0;
  for (int i = 0; i < n; ++i) {
    int pt = 0;
    for (int k = DEPTH; k >= 0; --k) {
      int id = (v[i] >> k) & 1;
      assert(pt < NN);
      if (trie[pt][id] == -1) {
        trie.emplace_back(BIN, -1);
        trie[pt][id] = trie.size() - 1;
      }
      ind[pt][id].push_back(i);
      pt = trie[pt][id];
      remax(pt_mx, pt);
    }
  }
  dfs();
  long long inv = 0;
  int x = 0;
  for (int d = DEPTH; d >= 0; --d) {
    inv += min(dp[d][0], dp[d][1]);
    if (dp[d][1] < dp[d][0]) {
      x += 1 << d;
    }
  }
  printf("%lld %d\n", inv, x);
}
int main(int argc, const char** argv) {
  solve();
  return 0;
}
