#include <bits/stdc++.h>
using namespace std;
mt19937 rnd(228);
const double EPS = 1e-7;
const double PI = acos(-1);
void solve();
signed main() {
  ios_base::sync_with_stdio(false);
  solve();
  exit(0);
}
void solve() {
  int n;
  cin >> n;
  vector<long long> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  int x = 0;
  long long ans = 0;
  vector<long long> b = a;
  for (long long i = 31; i >= 0; i--) {
    long long inv0 = 0, inv1 = 0, cnt0 = 0, cnt1 = 0;
    if ((b[0] & (1LL << i)) == 0)
      cnt0++;
    else
      cnt1++;
    for (int j = 1; j < n; j++) {
      if ((b[j] >> (i + 1)) == (b[j - 1] >> (i + 1))) {
        if ((b[j] & (1LL << i)) == 0) {
          inv0 += cnt1;
          cnt0++;
        } else {
          inv1 += cnt0;
          cnt1++;
        }
      } else {
        cnt0 = cnt1 = 0;
        if ((b[j] & (1LL << i)) == 0)
          cnt0++;
        else
          cnt1++;
      }
    }
    if (inv0 > inv1) {
      x += (1 << i);
      for (int j = 0; j < n; j++) {
        b[j] = (b[j] ^ (1LL << i));
      }
    }
    ans += min(inv0, inv1);
    vector<long long> c;
    int k = 0;
    if ((b[0] & (1LL << i)) == 0) {
      c.push_back(b[0]);
    }
    for (int j = 1; j < n; j++) {
      if ((b[j] >> (i + 1LL)) == (b[j - 1] >> (i + 1LL))) {
        if ((b[j] & (1LL << i)) == 0) {
          c.push_back(b[j]);
        }
      } else {
        while (k < j) {
          if ((b[k] & (1LL << i))) {
            c.push_back(b[k]);
          }
          k++;
        }
        if ((b[j] & (1LL << i)) == 0) {
          c.push_back(b[j]);
        }
      }
    }
    while (k < n) {
      if ((b[k] & (1LL << i))) {
        c.push_back(b[k]);
      }
      k++;
    }
    assert((int)c.size() == n);
    for (int j = 0; j < n; j++) {
      b[j] = c[j];
    }
  }
  cout << ans << ' ' << x;
}

### 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;
mt19937 rnd(228);
const double EPS = 1e-7;
const double PI = acos(-1);
void solve();
signed main() {
  ios_base::sync_with_stdio(false);
  solve();
  exit(0);
}
void solve() {
  int n;
  cin >> n;
  vector<long long> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  int x = 0;
  long long ans = 0;
  vector<long long> b = a;
  for (long long i = 31; i >= 0; i--) {
    long long inv0 = 0, inv1 = 0, cnt0 = 0, cnt1 = 0;
    if ((b[0] & (1LL << i)) == 0)
      cnt0++;
    else
      cnt1++;
    for (int j = 1; j < n; j++) {
      if ((b[j] >> (i + 1)) == (b[j - 1] >> (i + 1))) {
        if ((b[j] & (1LL << i)) == 0) {
          inv0 += cnt1;
          cnt0++;
        } else {
          inv1 += cnt0;
          cnt1++;
        }
      } else {
        cnt0 = cnt1 = 0;
        if ((b[j] & (1LL << i)) == 0)
          cnt0++;
        else
          cnt1++;
      }
    }
    if (inv0 > inv1) {
      x += (1 << i);
      for (int j = 0; j < n; j++) {
        b[j] = (b[j] ^ (1LL << i));
      }
    }
    ans += min(inv0, inv1);
    vector<long long> c;
    int k = 0;
    if ((b[0] & (1LL << i)) == 0) {
      c.push_back(b[0]);
    }
    for (int j = 1; j < n; j++) {
      if ((b[j] >> (i + 1LL)) == (b[j - 1] >> (i + 1LL))) {
        if ((b[j] & (1LL << i)) == 0) {
          c.push_back(b[j]);
        }
      } else {
        while (k < j) {
          if ((b[k] & (1LL << i))) {
            c.push_back(b[k]);
          }
          k++;
        }
        if ((b[j] & (1LL << i)) == 0) {
          c.push_back(b[j]);
        }
      }
    }
    while (k < n) {
      if ((b[k] & (1LL << i))) {
        c.push_back(b[k]);
      }
      k++;
    }
    assert((int)c.size() == n);
    for (int j = 0; j < n; j++) {
      b[j] = c[j];
    }
  }
  cout << ans << ' ' << x;
}

```