#include <bits/stdc++.h>
using namespace std;
const long long N = 4e6 + 100;
long long n, dp[30][2], t[2][N];
long long id = 1;
vector<long long> g[N];
long long ans = 0, x = 0;
void add(long long v, long long idx) {
  long long ver = 0;
  for (long long b = 29; b > -1; b--) {
    long long x = ((v >> b) & 1ll);
    if (!t[x][ver]) t[x][ver] = id++;
    ver = t[x][ver];
    g[ver].push_back(idx);
  }
}
void go(long long ver, long long b) {
  long long l = t[0][ver], r = t[1][ver];
  if (l) go(l, b - 1);
  if (r) go(r, b - 1);
  if (!l || !r) return;
  long long cnt = 0;
  long long i = 0, j = 0;
  while (i < g[l].size()) {
    while (j < g[r].size() && g[r][j] < g[l][i]) j++;
    cnt += j;
    i++;
  }
  dp[b][0] += cnt;
  dp[b][1] += g[l].size() * 1ll * g[r].size() - cnt;
  return;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    add(x, i);
  }
  go(0, 29);
  for (int b = 0; b < 30; b++) {
    ans += min(dp[b][0], dp[b][1]);
    if (dp[b][0] > dp[b][1]) x += (1ll << b);
  }
  cout << ans << " " << x;
}

### Prompt

In CPP, your task is to solve 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 long long N = 4e6 + 100;
long long n, dp[30][2], t[2][N];
long long id = 1;
vector<long long> g[N];
long long ans = 0, x = 0;
void add(long long v, long long idx) {
  long long ver = 0;
  for (long long b = 29; b > -1; b--) {
    long long x = ((v >> b) & 1ll);
    if (!t[x][ver]) t[x][ver] = id++;
    ver = t[x][ver];
    g[ver].push_back(idx);
  }
}
void go(long long ver, long long b) {
  long long l = t[0][ver], r = t[1][ver];
  if (l) go(l, b - 1);
  if (r) go(r, b - 1);
  if (!l || !r) return;
  long long cnt = 0;
  long long i = 0, j = 0;
  while (i < g[l].size()) {
    while (j < g[r].size() && g[r][j] < g[l][i]) j++;
    cnt += j;
    i++;
  }
  dp[b][0] += cnt;
  dp[b][1] += g[l].size() * 1ll * g[r].size() - cnt;
  return;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    add(x, i);
  }
  go(0, 29);
  for (int b = 0; b < 30; b++) {
    ans += min(dp[b][0], dp[b][1]);
    if (dp[b][0] > dp[b][1]) x += (1ll << b);
  }
  cout << ans << " " << x;
}

```