#include <bits/stdc++.h>
using namespace std;
const int maxn = 4e6 + 100;
int n, Trie[maxn][2], tot;
long long dp[maxn][2];
vector<int> g[maxn];
void add(int x, int p) {
  int u = 0;
  for (int i = 29; i >= 0; i--) {
    int bit = ((x >> i) & 1);
    if (!Trie[u][bit]) Trie[u][bit] = ++tot;
    u = Trie[u][bit];
    g[u].push_back(p);
  }
}
void dfs(int u, int dep) {
  if (Trie[u][0]) dfs(Trie[u][0], dep - 1);
  if (Trie[u][1]) dfs(Trie[u][1], dep - 1);
  if (!Trie[u][0] || !Trie[u][1]) return;
  long long sum = 0;
  int tt = 0;
  for (int i = 0; i < g[Trie[u][0]].size(); i++) {
    while (tt < g[Trie[u][1]].size() && g[Trie[u][1]][tt] < g[Trie[u][0]][i])
      tt++;
    sum += tt;
  }
  dp[dep][0] += sum;
  dp[dep][1] += (long long)g[Trie[u][0]].size() * g[Trie[u][1]].size() - sum;
}
void solve() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    add(x, i);
  }
  dfs(0, 29);
  long long ans = 0, u = 0;
  for (int i = 0; i <= 29; i++) {
    ans += min(dp[i][0], dp[i][1]);
    if (dp[i][1] < dp[i][0]) u += (1 << i);
  }
  printf("%lld %lld\n", ans, u);
}
int main() { solve(); }

### Prompt

Generate 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 = 4e6 + 100;
int n, Trie[maxn][2], tot;
long long dp[maxn][2];
vector<int> g[maxn];
void add(int x, int p) {
  int u = 0;
  for (int i = 29; i >= 0; i--) {
    int bit = ((x >> i) & 1);
    if (!Trie[u][bit]) Trie[u][bit] = ++tot;
    u = Trie[u][bit];
    g[u].push_back(p);
  }
}
void dfs(int u, int dep) {
  if (Trie[u][0]) dfs(Trie[u][0], dep - 1);
  if (Trie[u][1]) dfs(Trie[u][1], dep - 1);
  if (!Trie[u][0] || !Trie[u][1]) return;
  long long sum = 0;
  int tt = 0;
  for (int i = 0; i < g[Trie[u][0]].size(); i++) {
    while (tt < g[Trie[u][1]].size() && g[Trie[u][1]][tt] < g[Trie[u][0]][i])
      tt++;
    sum += tt;
  }
  dp[dep][0] += sum;
  dp[dep][1] += (long long)g[Trie[u][0]].size() * g[Trie[u][1]].size() - sum;
}
void solve() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    add(x, i);
  }
  dfs(0, 29);
  long long ans = 0, u = 0;
  for (int i = 0; i <= 29; i++) {
    ans += min(dp[i][0], dp[i][1]);
    if (dp[i][1] < dp[i][0]) u += (1 << i);
  }
  printf("%lld %lld\n", ans, u);
}
int main() { solve(); }

```