#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 3e5 + 69;
const long long LOGA = 31;
long long n;
long long trie[MAXN * 16][2], now;
vector<long long> g[MAXN * 16];
long long dp[LOGA][2];
void add(long long x) {
  long long v = 0;
  for (long long i = LOGA - 1; i >= 0; i--) {
    long long bit = (x >> i) & 1;
    if (!trie[v][bit]) trie[v][bit] = ++now;
    g[v].push_back(x);
    v = trie[v][bit];
    if (i == 0) g[v].push_back(x);
  }
}
void dfs(long long v, long long i) {
  if (i < 0) return;
  long long l = trie[v][0], r = trie[v][1];
  if (l) dfs(l, i - 1);
  if (r) dfs(r, i - 1);
  if (!l || !r) return;
  long long cnt = 0;
  long long cnt1 = 0;
  for (long long u : g[v]) {
    if ((u >> i) & 1)
      cnt1++;
    else
      cnt += cnt1;
  }
  dp[i][0] += cnt;
  dp[i][1] += 1ll * g[l].size() * g[r].size() - cnt;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 1, x; i <= n; i++) {
    cin >> x;
    add(x);
  }
  dfs(0, LOGA - 1);
  long long ans1 = 0, ans2 = 0;
  for (long long i = LOGA - 1; i >= 0; i--) {
    ans1 += min(dp[i][0], dp[i][1]);
    if (dp[i][0] > dp[i][1]) ans2 |= (1ll << i);
  }
  cout << ans1 << ' ' << ans2;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 long long MAXN = 3e5 + 69;
const long long LOGA = 31;
long long n;
long long trie[MAXN * 16][2], now;
vector<long long> g[MAXN * 16];
long long dp[LOGA][2];
void add(long long x) {
  long long v = 0;
  for (long long i = LOGA - 1; i >= 0; i--) {
    long long bit = (x >> i) & 1;
    if (!trie[v][bit]) trie[v][bit] = ++now;
    g[v].push_back(x);
    v = trie[v][bit];
    if (i == 0) g[v].push_back(x);
  }
}
void dfs(long long v, long long i) {
  if (i < 0) return;
  long long l = trie[v][0], r = trie[v][1];
  if (l) dfs(l, i - 1);
  if (r) dfs(r, i - 1);
  if (!l || !r) return;
  long long cnt = 0;
  long long cnt1 = 0;
  for (long long u : g[v]) {
    if ((u >> i) & 1)
      cnt1++;
    else
      cnt += cnt1;
  }
  dp[i][0] += cnt;
  dp[i][1] += 1ll * g[l].size() * g[r].size() - cnt;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 1, x; i <= n; i++) {
    cin >> x;
    add(x);
  }
  dfs(0, LOGA - 1);
  long long ans1 = 0, ans2 = 0;
  for (long long i = LOGA - 1; i >= 0; i--) {
    ans1 += min(dp[i][0], dp[i][1]);
    if (dp[i][0] > dp[i][1]) ans2 |= (1ll << i);
  }
  cout << ans1 << ' ' << ans2;
}

```