#include <bits/stdc++.h>
using namespace std;
vector<pair<long long int, long long int>> lr;
vector<long long int> dp, dp1;
vector<long long int> sum;
vector<pair<long long int, long long int>> arr;
int bi[300001][32];
void settup() {
  vector<pair<long long int, long long int>> bau = arr;
  for (int j = 0; j < 32; j++)
    for (int i = 0; i < bau.size(); i++) {
      bi[i][j] = bau[i].first % 2;
      bau[i].first = bau[i].first / 2;
    }
}
int InvSe(int l, int r, int length);
int main() {
  int n;
  cin >> n;
  arr.push_back(make_pair(0, 0));
  sum.assign(60, 0);
  dp.assign(60, 0);
  dp1.assign(60, 0);
  for (int i = 0; i < n; i++) {
    int y;
    cin >> y;
    arr.push_back(make_pair(y, i + 1));
  }
  sort(arr.begin() + 1, arr.end());
  settup();
  InvSe(1, n, 31);
  long long int ans = 0, u = 0;
  for (int i = 31; i >= 0; i--) {
    ans += min(sum[i] - dp[i], dp[i]);
    if (dp[i] > sum[i] - dp[i]) u += pow(2, i);
  }
  cout << ans << " " << u << endl;
  getchar();
  getchar();
  return 0;
}
int InvSe(int l, int r, int length) {
  if (l >= r || length == -1) {
    return 0;
  } else {
    long long int m = (l + r) / 2;
    long long int x, y = 0, siz = 0;
    for (int i = l; i <= r; i++) {
      if (bi[i][length] == 0) siz++;
    }
    if (siz != r - l + 1 && siz != 0) {
      x = InvSe(l, l + siz - 1, length - 1);
      y = InvSe(l + siz, r, length - 1);
      vector<long long int> te1, te2;
      for (int i = l; i < l + siz; i++) te1.push_back(arr[i].second);
      for (int i = l + siz; i <= r; i++) te2.push_back(arr[i].second);
      int i = 0, j = 0, k = l;
      int n2 = r - l - siz + 1;
      long long int n1 = siz;
      long long int c = 0, c1 = 0;
      while (i < n1 && j < n2) {
        if (te1[i] < te2[j]) {
          arr[k].second = te1[i];
          i++;
        } else {
          c += n1 - i;
          arr[k].second = te2[j];
          j++;
        }
        k++;
      }
      while (i < n1) {
        arr[k].second = te1[i];
        i++;
        k++;
      }
      while (j < n2) {
        arr[k].second = te2[j];
        j++;
        k++;
      }
      sum[length] += siz * (r - l + 1 - siz);
      dp[length] += c;
      return c;
    }
    return InvSe(l, r, length - 1);
  }
}

### 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;
vector<pair<long long int, long long int>> lr;
vector<long long int> dp, dp1;
vector<long long int> sum;
vector<pair<long long int, long long int>> arr;
int bi[300001][32];
void settup() {
  vector<pair<long long int, long long int>> bau = arr;
  for (int j = 0; j < 32; j++)
    for (int i = 0; i < bau.size(); i++) {
      bi[i][j] = bau[i].first % 2;
      bau[i].first = bau[i].first / 2;
    }
}
int InvSe(int l, int r, int length);
int main() {
  int n;
  cin >> n;
  arr.push_back(make_pair(0, 0));
  sum.assign(60, 0);
  dp.assign(60, 0);
  dp1.assign(60, 0);
  for (int i = 0; i < n; i++) {
    int y;
    cin >> y;
    arr.push_back(make_pair(y, i + 1));
  }
  sort(arr.begin() + 1, arr.end());
  settup();
  InvSe(1, n, 31);
  long long int ans = 0, u = 0;
  for (int i = 31; i >= 0; i--) {
    ans += min(sum[i] - dp[i], dp[i]);
    if (dp[i] > sum[i] - dp[i]) u += pow(2, i);
  }
  cout << ans << " " << u << endl;
  getchar();
  getchar();
  return 0;
}
int InvSe(int l, int r, int length) {
  if (l >= r || length == -1) {
    return 0;
  } else {
    long long int m = (l + r) / 2;
    long long int x, y = 0, siz = 0;
    for (int i = l; i <= r; i++) {
      if (bi[i][length] == 0) siz++;
    }
    if (siz != r - l + 1 && siz != 0) {
      x = InvSe(l, l + siz - 1, length - 1);
      y = InvSe(l + siz, r, length - 1);
      vector<long long int> te1, te2;
      for (int i = l; i < l + siz; i++) te1.push_back(arr[i].second);
      for (int i = l + siz; i <= r; i++) te2.push_back(arr[i].second);
      int i = 0, j = 0, k = l;
      int n2 = r - l - siz + 1;
      long long int n1 = siz;
      long long int c = 0, c1 = 0;
      while (i < n1 && j < n2) {
        if (te1[i] < te2[j]) {
          arr[k].second = te1[i];
          i++;
        } else {
          c += n1 - i;
          arr[k].second = te2[j];
          j++;
        }
        k++;
      }
      while (i < n1) {
        arr[k].second = te1[i];
        i++;
        k++;
      }
      while (j < n2) {
        arr[k].second = te2[j];
        j++;
        k++;
      }
      sum[length] += siz * (r - l + 1 - siz);
      dp[length] += c;
      return c;
    }
    return InvSe(l, r, length - 1);
  }
}

```