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;
const int INF = INT_MAX / 4, SINF = 0x3f;
const int dx[] = {0, 0, -1, 1}, dy[] = {-1, 1, 0, 0};
const int maxn = 3e5 + 10;
int mo_num = 1e9 + 7;
int n, t, case_of_T;
long long f[33][3], ans, m;
long long st[maxn << 4][3], cnt = 0;
unordered_map<int, vector<int> > h;
void Insert(int x, int pos) {
  int now = 0;
  for (int i = 30; i >= 0; i--) {
    int val = 1LL * x >> i * 1LL & 1LL;
    if (!st[now][val]) st[now][val] = ++cnt;
    h[now = st[now][val]].push_back(pos);
  }
}
void deal(int now, int pos) {
  int z = st[now][0], y = st[now][1];
  if (!h[z].empty() && !h[y].empty()) {
    sort(h[z].begin(), h[z].end());
    sort(h[y].begin(), h[y].end());
    long long zuo = 0, you = 0, sz = h[z].size(), sy = h[y].size();
    long long num = 0;
    while (zuo < sz && you < sy) {
      if (h[z][zuo] > h[y][you])
        you++, num += 1LL * sz - 1LL * zuo;
      else
        zuo++;
    }
    f[pos][0] += 1LL * num;
    f[pos][1] += 1LL * sz * sy - 1LL * num;
  }
  if (z && pos) deal(z, pos - 1);
  if (y && pos) deal(y, pos - 1);
}
void solve() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    int x;
    cin >> x;
    Insert(x, i);
  }
  deal(0, 30);
  for (int i = 30; i >= 0; i--) {
    if (f[i][0] <= f[i][1]) {
      ans += 1LL * f[i][0];
    } else {
      ans += 1LL * f[i][1];
      m |= (1LL << i);
    }
  }
  cout << ans << " " << m << '\n';
  return;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  while (T--) solve();
  return 0;
}
