#include <bits/stdc++.h>
using namespace std;
const int maxn = 5e6 + 100;
int trie[maxn][2], id = 1, n;
long long dp[30][2];
vector<int> g[maxn];
void add(int a, int ind) {
  int c = 0;
  for (int i = 29; i >= 0; i--) {
    int bit = (a >> i) & 1;
    if (!trie[c][bit]) {
      trie[c][bit] = id++;
    }
    c = trie[c][bit];
    g[c].push_back(ind);
  }
}
void go(int c = 0, int d = 29) {
  int l = trie[c][0], r = trie[c][1];
  if (l) {
    go(l, d - 1);
  }
  if (r) {
    go(r, d - 1);
  }
  if (!l || !r) {
    return;
  }
  long long res = 0ll;
  int ptr = 0;
  for (int x : g[l]) {
    while (ptr < g[r].size() && g[r][ptr] < x) {
      ptr++;
    }
    res += ptr;
  }
  dp[d][0] += 0ll + res;
  dp[d][1] += 1ll * g[r].size() * g[l].size() - res;
}
void solve() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    add(x, i);
  }
  go();
  long long res = 0ll;
  int ans = 0;
  for (int i = 0; i < 30; i++) {
    res += min(dp[i][0], dp[i][1]);
    if (dp[i][1] < dp[i][0]) {
      ans |= (1 << i);
    }
  }
  cout << res << " " << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Please formulate 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;
const int maxn = 5e6 + 100;
int trie[maxn][2], id = 1, n;
long long dp[30][2];
vector<int> g[maxn];
void add(int a, int ind) {
  int c = 0;
  for (int i = 29; i >= 0; i--) {
    int bit = (a >> i) & 1;
    if (!trie[c][bit]) {
      trie[c][bit] = id++;
    }
    c = trie[c][bit];
    g[c].push_back(ind);
  }
}
void go(int c = 0, int d = 29) {
  int l = trie[c][0], r = trie[c][1];
  if (l) {
    go(l, d - 1);
  }
  if (r) {
    go(r, d - 1);
  }
  if (!l || !r) {
    return;
  }
  long long res = 0ll;
  int ptr = 0;
  for (int x : g[l]) {
    while (ptr < g[r].size() && g[r][ptr] < x) {
      ptr++;
    }
    res += ptr;
  }
  dp[d][0] += 0ll + res;
  dp[d][1] += 1ll * g[r].size() * g[l].size() - res;
}
void solve() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    add(x, i);
  }
  go();
  long long res = 0ll;
  int ans = 0;
  for (int i = 0; i < 30; i++) {
    res += min(dp[i][0], dp[i][1]);
    if (dp[i][1] < dp[i][0]) {
      ans |= (1 << i);
    }
  }
  cout << res << " " << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```