#include <bits/stdc++.h>
using namespace std;
pair<long long, int> getBit(vector<vector<int>> &gs, int ib) {
  long long ni[2] = {0};
  for (int b = 0; b < 2; b++) {
    for (auto &v : gs) {
      int n1 = 0;
      for (auto &x : v)
        if (((x >> ib) ^ b) & 1)
          n1++;
        else
          ni[b] += n1;
    }
  }
  if (ni[0] <= ni[1])
    return {ni[0], 0};
  else
    return {ni[1], 1};
}
int main() {
  int n;
  while (scanf("%d", &n) == 1) {
    vector<int> a(n);
    for (auto &x : a) cin >> x;
    long long ni = 0;
    int x = 0;
    vector<vector<int>> gs = {a};
    for (int ib = 30; ib >= 0; ib--) {
      auto [tni, b] = getBit(gs, ib);
      x |= b << ib;
      ni += tni;
      vector<vector<int>> ngs;
      for (auto &v : gs) {
        vector<int> b[2];
        for (auto &x : v) b[x >> ib & 1].push_back(x);
        for (int i = 0; i < 2; i++)
          if (!b[i].empty()) ngs.push_back(b[i]);
      }
      gs = ngs;
    }
    printf("%lld %d\n", ni, x);
  }
}

### Prompt

Please create a solution in Cpp 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;
pair<long long, int> getBit(vector<vector<int>> &gs, int ib) {
  long long ni[2] = {0};
  for (int b = 0; b < 2; b++) {
    for (auto &v : gs) {
      int n1 = 0;
      for (auto &x : v)
        if (((x >> ib) ^ b) & 1)
          n1++;
        else
          ni[b] += n1;
    }
  }
  if (ni[0] <= ni[1])
    return {ni[0], 0};
  else
    return {ni[1], 1};
}
int main() {
  int n;
  while (scanf("%d", &n) == 1) {
    vector<int> a(n);
    for (auto &x : a) cin >> x;
    long long ni = 0;
    int x = 0;
    vector<vector<int>> gs = {a};
    for (int ib = 30; ib >= 0; ib--) {
      auto [tni, b] = getBit(gs, ib);
      x |= b << ib;
      ni += tni;
      vector<vector<int>> ngs;
      for (auto &v : gs) {
        vector<int> b[2];
        for (auto &x : v) b[x >> ib & 1].push_back(x);
        for (int i = 0; i < 2; i++)
          if (!b[i].empty()) ngs.push_back(b[i]);
      }
      gs = ngs;
    }
    printf("%lld %d\n", ni, x);
  }
}

```