#include <bits/stdc++.h>
using namespace std;
template <class Type>
long long quick_pow(Type x, Type y, Type Mod) {
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x % Mod;
    x = x * x % Mod, y >>= 1;
  }
  return res;
}
template <class Type>
long long inv(Type x, Type Mod) {
  return quick(x, Mod - 2, Mod);
}
namespace IO {
template <class T>
inline void read(T &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
  x *= f;
}
};  // namespace IO
using namespace IO;
const int N = 300005;
int n, bin[50];
int t[N][2];
long long ans[50][2];
int lowbit(int x) { return x & -x; }
void add(int x, int y, int flag) {
  while (x < N) t[x][flag] += y, x += lowbit(x);
}
int query(int x, int flag) {
  int res = 0;
  while (x > 0) {
    res += t[x][flag], x -= lowbit(x);
  }
  return res;
}
struct S {
  int num, id;
} s[N];
bool cmp(S x, S y) {
  if (x.num == y.num)
    return x.id < y.id;
  else
    return x.num < y.num;
}
void DFS(int L, int R, int pos) {
  if (pos < 0) return;
  if (L > R) return;
  int num0 = 0;
  for (int i = (L); i <= (R); ++i) {
    if (s[i].num & bin[pos])
      add(s[i].id, 1, 1);
    else
      add(s[i].id, 1, 0), num0++;
  }
  for (int i = (L); i <= (R); ++i) {
    if (s[i].num & bin[pos])
      ans[pos][0] += (long long)query(s[i].id - 1, 0);
    else
      ans[pos][1] += (long long)query(s[i].id - 1, 1);
  }
  for (int i = (L); i <= (R); ++i) {
    if (s[i].num & bin[pos])
      add(s[i].id, -1, 1);
    else
      add(s[i].id, -1, 0);
  }
  sort(s + L, s + R + 1, cmp);
  for (int i = (L); i <= (R); ++i)
    if (s[i].num & bin[pos]) s[i].num -= bin[pos];
  DFS(L, L + num0 - 1, pos - 1);
  DFS(L + num0, R, pos - 1);
}
int Ansnum;
long long Ans;
int main() {
  read(n);
  for (int i = (1); i <= (n); ++i) read(s[i].num), s[i].id = i;
  bin[0] = 1;
  for (int i = (1); i <= (30); ++i) bin[i] = bin[i - 1] * 2;
  DFS(1, n, 30);
  for (int i = (30); i >= (0); --i) {
    if (ans[i][0] < ans[i][1]) Ansnum += bin[i];
    Ans += min(ans[i][0], ans[i][1]);
  }
  printf("%lld %d\n", Ans, Ansnum);
  return 0;
}

### Prompt

Your challenge is to write 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 <class Type>
long long quick_pow(Type x, Type y, Type Mod) {
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x % Mod;
    x = x * x % Mod, y >>= 1;
  }
  return res;
}
template <class Type>
long long inv(Type x, Type Mod) {
  return quick(x, Mod - 2, Mod);
}
namespace IO {
template <class T>
inline void read(T &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
  x *= f;
}
};  // namespace IO
using namespace IO;
const int N = 300005;
int n, bin[50];
int t[N][2];
long long ans[50][2];
int lowbit(int x) { return x & -x; }
void add(int x, int y, int flag) {
  while (x < N) t[x][flag] += y, x += lowbit(x);
}
int query(int x, int flag) {
  int res = 0;
  while (x > 0) {
    res += t[x][flag], x -= lowbit(x);
  }
  return res;
}
struct S {
  int num, id;
} s[N];
bool cmp(S x, S y) {
  if (x.num == y.num)
    return x.id < y.id;
  else
    return x.num < y.num;
}
void DFS(int L, int R, int pos) {
  if (pos < 0) return;
  if (L > R) return;
  int num0 = 0;
  for (int i = (L); i <= (R); ++i) {
    if (s[i].num & bin[pos])
      add(s[i].id, 1, 1);
    else
      add(s[i].id, 1, 0), num0++;
  }
  for (int i = (L); i <= (R); ++i) {
    if (s[i].num & bin[pos])
      ans[pos][0] += (long long)query(s[i].id - 1, 0);
    else
      ans[pos][1] += (long long)query(s[i].id - 1, 1);
  }
  for (int i = (L); i <= (R); ++i) {
    if (s[i].num & bin[pos])
      add(s[i].id, -1, 1);
    else
      add(s[i].id, -1, 0);
  }
  sort(s + L, s + R + 1, cmp);
  for (int i = (L); i <= (R); ++i)
    if (s[i].num & bin[pos]) s[i].num -= bin[pos];
  DFS(L, L + num0 - 1, pos - 1);
  DFS(L + num0, R, pos - 1);
}
int Ansnum;
long long Ans;
int main() {
  read(n);
  for (int i = (1); i <= (n); ++i) read(s[i].num), s[i].id = i;
  bin[0] = 1;
  for (int i = (1); i <= (30); ++i) bin[i] = bin[i - 1] * 2;
  DFS(1, n, 30);
  for (int i = (30); i >= (0); --i) {
    if (ans[i][0] < ans[i][1]) Ansnum += bin[i];
    Ans += min(ans[i][0], ans[i][1]);
  }
  printf("%lld %d\n", Ans, Ansnum);
  return 0;
}

```