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;
using pii = pair<int, int>;
using lli = long long int;
using iii = tuple<int, int, int>;
using iiii = tuple<int, int, int, int>;
using ulli = unsigned long long int;
using ui = unsigned int;
using Bit = bitset<30>;
const int maxBit = 30;
const int maxN = 3e5;
const int block = 600;
const int cBlocks = maxN / block;
const int C = 1e3;
const int maxM = 2e5;
const int maxl = 18;
const int inf = 0x3f3f3f3f;
const lli INF = 0x3f3f3f3f3f3f3f3f;
const char sp = ' ';
const char dl = '\n';
const int mod = 998244353;
int nxt[maxN * maxBit + 5][2], numNode;
vector<int> List[maxN * maxBit + 5];
lli f[maxBit][2];
void Insert(const Bit& a, const int& idx) {
  int cur = 0;
  for (int i = maxBit - 1; i >= 0; i--) {
    if (!nxt[cur][a[i]]) nxt[cur][a[i]] = ++numNode;
    cur = nxt[cur][a[i]];
    List[cur].push_back(idx);
  }
  return;
}
void dfs(int cur, int bit) {
  int l = nxt[cur][0];
  int r = nxt[cur][1];
  if (l) dfs(l, bit - 1);
  if (r) dfs(r, bit - 1);
  if (!l || !r) return;
  lli res = 0;
  int j = 0;
  for (int x : List[l]) {
    while (j < ((int)List[r].size()) && List[r][j] < x) j++;
    res += j;
  }
  f[bit][0] += res;
  f[bit][1] += 1ll * ((int)List[l].size()) * ((int)List[r].size()) - res;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int test = 1;
  while (test--) {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      Insert(Bit(a[i]), i);
    }
    dfs(0, 29);
    lli res = 0, cnt = 0;
    for (int i = maxBit - 1; i >= 0; i--) {
      cnt += min(f[i][0], f[i][1]);
      if (f[i][1] < f[i][0]) res += (1 << i);
    }
    cout << cnt << ' ' << res << '\n';
  }
  return 0;
}
