#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
const double pi = acos(-1);
const int inf = 2e9;
const int maxn = 3e5 + 15;
int read() {
  char ch = getchar();
  int x = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int a[maxn];
int b[maxn];
int c[maxn];
int p[maxn];
int d;
int n;
void add(int x) {
  for (int i = x; d >= i; i += i & -i) {
    c[i]++;
  }
}
int query(int x) {
  int ans = 0;
  for (long long i = x; i; i -= i & -i) {
    ans += c[i];
  }
  return ans;
}
long long ans;
int pos[maxn];
void solve(int t, int num, long long preans1) {
  if (t == -1) {
    printf("%lld %d\n", ans, num);
    return;
  }
  long long ans2 = 0;
  int val = num + (1 << t);
  for (int i = 1; n >= i; ++i) {
    b[i] = a[i] ^ val;
    p[i] = b[i];
    c[i] = 0;
  }
  sort(p + 1, p + n + 1);
  d = unique(p + 1, p + n + 1) - p - 1;
  int tmp = n % 2;
  for (int i = 1; tmp >= i; ++i) {
    pos[i] = lower_bound(p + 1, p + d + 1, b[i]) - p;
  }
  for (int i = 1 + tmp; n >= i; i += 2) {
    pos[i] = lower_bound(p + 1, p + d + 1, b[i]) - p;
    pos[i + 1] = lower_bound(p + 1, p + d + 1, b[i + 1]) - p;
  }
  for (int i = n; i >= 1; --i) {
    if (pos[i] - 1 >= 1) ans2 += query(pos[i] - 1);
    add(pos[i]);
    if (ans2 >= preans1) break;
  }
  ans = min(preans1, ans2);
  if (preans1 <= ans2) {
    solve(t - 1, num, preans1);
  } else {
    solve(t - 1, val, ans2);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; n >= i; ++i) {
    a[i] = read();
  }
  for (int i = 1; n >= i; ++i) {
    b[i] = a[i];
    p[i] = b[i];
    c[i] = 0;
  }
  long long ans1 = 0;
  sort(p + 1, p + n + 1);
  d = unique(p + 1, p + n + 1) - p - 1;
  for (int i = n; i >= 1; --i) {
    int pos = lower_bound(p + 1, p + d + 1, b[i]) - p;
    if (pos - 1 >= 1) ans1 += query(pos - 1);
    add(pos);
  }
  solve(29, 0, ans1);
  return 0;
}

### Prompt

Your task is to create 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 mod = 998244353;
const double pi = acos(-1);
const int inf = 2e9;
const int maxn = 3e5 + 15;
int read() {
  char ch = getchar();
  int x = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int a[maxn];
int b[maxn];
int c[maxn];
int p[maxn];
int d;
int n;
void add(int x) {
  for (int i = x; d >= i; i += i & -i) {
    c[i]++;
  }
}
int query(int x) {
  int ans = 0;
  for (long long i = x; i; i -= i & -i) {
    ans += c[i];
  }
  return ans;
}
long long ans;
int pos[maxn];
void solve(int t, int num, long long preans1) {
  if (t == -1) {
    printf("%lld %d\n", ans, num);
    return;
  }
  long long ans2 = 0;
  int val = num + (1 << t);
  for (int i = 1; n >= i; ++i) {
    b[i] = a[i] ^ val;
    p[i] = b[i];
    c[i] = 0;
  }
  sort(p + 1, p + n + 1);
  d = unique(p + 1, p + n + 1) - p - 1;
  int tmp = n % 2;
  for (int i = 1; tmp >= i; ++i) {
    pos[i] = lower_bound(p + 1, p + d + 1, b[i]) - p;
  }
  for (int i = 1 + tmp; n >= i; i += 2) {
    pos[i] = lower_bound(p + 1, p + d + 1, b[i]) - p;
    pos[i + 1] = lower_bound(p + 1, p + d + 1, b[i + 1]) - p;
  }
  for (int i = n; i >= 1; --i) {
    if (pos[i] - 1 >= 1) ans2 += query(pos[i] - 1);
    add(pos[i]);
    if (ans2 >= preans1) break;
  }
  ans = min(preans1, ans2);
  if (preans1 <= ans2) {
    solve(t - 1, num, preans1);
  } else {
    solve(t - 1, val, ans2);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; n >= i; ++i) {
    a[i] = read();
  }
  for (int i = 1; n >= i; ++i) {
    b[i] = a[i];
    p[i] = b[i];
    c[i] = 0;
  }
  long long ans1 = 0;
  sort(p + 1, p + n + 1);
  d = unique(p + 1, p + n + 1) - p - 1;
  for (int i = n; i >= 1; --i) {
    int pos = lower_bound(p + 1, p + d + 1, b[i]) - p;
    if (pos - 1 >= 1) ans1 += query(pos - 1);
    add(pos);
  }
  solve(29, 0, ans1);
  return 0;
}

```