#include <bits/stdc++.h>
using namespace std;
inline long long getint() {
  char c = getchar();
  long long ret = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') ret = ret * 10 + c - '0', c = getchar();
  return ret * f;
}
const int maxn = 3e5 + 10;
vector<int> b[2][maxn];
int x = 0, y = 1, n, tot;
inline long long solve(int bin, int d) {
  long long ret = 0;
  for (int i = 1; i <= tot; i++) {
    int now = 0;
    for (int j = 0; j < b[x][i].size(); j++)
      if (b[x][i][j] >> bin & 1 ^ d)
        now++;
      else
        ret += now;
  }
  return ret;
}
inline void split(int bin) {
  int pos = 0;
  for (int i = 1; i <= tot; i++) {
    ++pos;
    int cnt = 0;
    for (auto j : b[x][i]) cnt |= (j >> bin & 1) + 1;
    for (auto j : b[x][i]) {
      if (j >> bin & 1) {
        if (cnt == 3)
          b[y][pos + 1].push_back(j);
        else
          b[y][pos].push_back(j);
      } else
        b[y][pos].push_back(j);
    }
    if (cnt == 3) pos++;
  }
  for (int i = 1; i <= tot; i++) b[x][i].clear();
  tot = pos;
  x ^= 1;
  y ^= 1;
}
int a[maxn];
int main() {
  n = getint();
  tot = 1;
  for (int i = 1; i <= n; i++) b[0][1].push_back(getint());
  long long ans1 = 0;
  int ans2 = 0;
  for (int i = 30; i >= 0; i--) {
    long long tmp0 = solve(i, 0), tmp1 = solve(i, 1);
    if (tmp1 < tmp0)
      ans2 |= 1 << i, ans1 += tmp1;
    else
      ans1 += tmp0;
    split(i);
  }
  printf("%lld %d\n", ans1, ans2);
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
inline long long getint() {
  char c = getchar();
  long long ret = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') ret = ret * 10 + c - '0', c = getchar();
  return ret * f;
}
const int maxn = 3e5 + 10;
vector<int> b[2][maxn];
int x = 0, y = 1, n, tot;
inline long long solve(int bin, int d) {
  long long ret = 0;
  for (int i = 1; i <= tot; i++) {
    int now = 0;
    for (int j = 0; j < b[x][i].size(); j++)
      if (b[x][i][j] >> bin & 1 ^ d)
        now++;
      else
        ret += now;
  }
  return ret;
}
inline void split(int bin) {
  int pos = 0;
  for (int i = 1; i <= tot; i++) {
    ++pos;
    int cnt = 0;
    for (auto j : b[x][i]) cnt |= (j >> bin & 1) + 1;
    for (auto j : b[x][i]) {
      if (j >> bin & 1) {
        if (cnt == 3)
          b[y][pos + 1].push_back(j);
        else
          b[y][pos].push_back(j);
      } else
        b[y][pos].push_back(j);
    }
    if (cnt == 3) pos++;
  }
  for (int i = 1; i <= tot; i++) b[x][i].clear();
  tot = pos;
  x ^= 1;
  y ^= 1;
}
int a[maxn];
int main() {
  n = getint();
  tot = 1;
  for (int i = 1; i <= n; i++) b[0][1].push_back(getint());
  long long ans1 = 0;
  int ans2 = 0;
  for (int i = 30; i >= 0; i--) {
    long long tmp0 = solve(i, 0), tmp1 = solve(i, 1);
    if (tmp1 < tmp0)
      ans2 |= 1 << i, ans1 += tmp1;
    else
      ans1 += tmp0;
    split(i);
  }
  printf("%lld %d\n", ans1, ans2);
  return 0;
}

```