#include <bits/stdc++.h>
using namespace std;
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  x *= f;
}
inline void write(long long x) {
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
const int N = 300005, V = N * 30;
int n, a[N];
long long ans;
int ansv, ALL;
int id[N];
inline bool cmp(int x, int y) { return (a[x] & ALL) < (a[y] & ALL); }
long long s0, s1;
int p0[N], l0, p1[N], l1;
struct BIT {
  int d[N];
  inline void add(int x) {
    while (x <= n) ++d[x], x += x & -x;
  }
  inline int ask(int x) {
    static int r;
    r = 0;
    while (x) r += d[x], x -= x & -x;
    return r;
  }
} T;
inline void workk(int L, int R, int d) {
  int i;
  long long sum = 0;
  ALL ^= 1 << d;
  l0 = l1 = 0;
  for (i = L; i <= R; ++i)
    if (a[id[i]] >> d & 1)
      p1[++l1] = id[i];
    else
      p0[++l0] = id[i];
  sort(p1 + 1, p1 + l1 + 1);
  sort(p0 + 1, p0 + l0 + 1);
  int cnt = 0;
  for (i = 1; i <= l0; ++i) {
    while (cnt + 1 <= l1 && p1[cnt + 1] <= p0[i]) ++cnt;
    sum += cnt;
  }
  s0 += sum, s1 += 1ll * l0 * l1 - sum;
  ALL ^= 1 << d;
}
inline void work(int b) {
  int i;
  s0 = s1 = 0;
  ALL = (1 << 30) - 1;
  for (i = 0; i <= b; ++i) ALL ^= 1 << i;
  for (i = 1; i <= n; ++i) id[i] = i;
  sort(id + 1, id + n + 1, cmp);
  int l = 1;
  for (i = 2; i <= n; ++i) {
    if ((a[id[i]] & ALL) != (a[id[i - 1]] & ALL)) workk(l, i - 1, b), l = i;
  }
  workk(l, n, b);
  if (s0 <= s1)
    ans += s0;
  else
    ans += s1, ansv |= 1 << b;
}
int main() {
  int i;
  read(n);
  for (i = 1; i <= n; ++i) read(a[i]);
  for (i = 0; i < 30; ++i) work(i);
  cout << ans << ' ' << ansv << '\n';
  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;
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  x *= f;
}
inline void write(long long x) {
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
const int N = 300005, V = N * 30;
int n, a[N];
long long ans;
int ansv, ALL;
int id[N];
inline bool cmp(int x, int y) { return (a[x] & ALL) < (a[y] & ALL); }
long long s0, s1;
int p0[N], l0, p1[N], l1;
struct BIT {
  int d[N];
  inline void add(int x) {
    while (x <= n) ++d[x], x += x & -x;
  }
  inline int ask(int x) {
    static int r;
    r = 0;
    while (x) r += d[x], x -= x & -x;
    return r;
  }
} T;
inline void workk(int L, int R, int d) {
  int i;
  long long sum = 0;
  ALL ^= 1 << d;
  l0 = l1 = 0;
  for (i = L; i <= R; ++i)
    if (a[id[i]] >> d & 1)
      p1[++l1] = id[i];
    else
      p0[++l0] = id[i];
  sort(p1 + 1, p1 + l1 + 1);
  sort(p0 + 1, p0 + l0 + 1);
  int cnt = 0;
  for (i = 1; i <= l0; ++i) {
    while (cnt + 1 <= l1 && p1[cnt + 1] <= p0[i]) ++cnt;
    sum += cnt;
  }
  s0 += sum, s1 += 1ll * l0 * l1 - sum;
  ALL ^= 1 << d;
}
inline void work(int b) {
  int i;
  s0 = s1 = 0;
  ALL = (1 << 30) - 1;
  for (i = 0; i <= b; ++i) ALL ^= 1 << i;
  for (i = 1; i <= n; ++i) id[i] = i;
  sort(id + 1, id + n + 1, cmp);
  int l = 1;
  for (i = 2; i <= n; ++i) {
    if ((a[id[i]] & ALL) != (a[id[i - 1]] & ALL)) workk(l, i - 1, b), l = i;
  }
  workk(l, n, b);
  if (s0 <= s1)
    ans += s0;
  else
    ans += s1, ansv |= 1 << b;
}
int main() {
  int i;
  read(n);
  for (i = 1; i <= n; ++i) read(a[i]);
  for (i = 0; i < 30; ++i) work(i);
  cout << ans << ' ' << ansv << '\n';
  return 0;
}

```