#include <bits/stdc++.h>
using namespace std;
long long getinv(vector<int>& val) {
  long long res = 0;
  long long cntone = 0;
  for (int i = 0; i < val.size(); ++i) {
    if (val[i] == 0) {
      res += cntone;
    } else {
      ++cntone;
    }
  }
  return res;
}
long long getinv(vector<vector<int> >& cur, int x) {
  long long res = 0;
  for (int i = 0; i < cur.size(); ++i) {
    vector<int> nv = cur[i];
    if (x == 1) {
      for (int j = 0; j < nv.size(); ++j) {
        nv[j] = 1 - nv[j];
      }
    }
    res += getinv(nv);
  }
  return res;
}
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  long long ans = 0;
  long long val = 0;
  vector<vector<int> > cur;
  cur.push_back(a);
  for (int ps = 29; ps >= 0; --ps) {
    vector<vector<int> > nv;
    for (int i = 0; i < cur.size(); ++i) {
      vector<int> ncur(cur[i].size(), 0);
      for (int j = 0; j < cur[i].size(); ++j) {
        ncur[j] = (bool)(cur[i][j] & (1 << ps));
      }
      nv.push_back(ncur);
    }
    long long rs1 = getinv(nv, 0);
    long long rs2 = getinv(nv, 1);
    if (rs2 < rs1) {
      ans += rs2;
      val += (1 << ps);
    } else {
      ans += rs1;
    }
    vector<vector<int> > nvi;
    for (int i = 0; i < cur.size(); ++i) {
      vector<int> nv1;
      vector<int> nv2;
      for (int j = 0; j < cur[i].size(); ++j) {
        if (cur[i][j] & (1 << ps)) {
          nv1.push_back(cur[i][j]);
        } else {
          nv2.push_back(cur[i][j]);
        }
      }
      if (!nv1.empty()) {
        nvi.push_back(nv1);
      }
      if (!nv2.empty()) {
        nvi.push_back(nv2);
      }
    }
    cur = nvi;
  }
  cout << ans << " " << val << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  t = 1;
  for (int tt = 0; tt < t; ++tt) {
    solve();
  }
  return 0;
}

### 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;
long long getinv(vector<int>& val) {
  long long res = 0;
  long long cntone = 0;
  for (int i = 0; i < val.size(); ++i) {
    if (val[i] == 0) {
      res += cntone;
    } else {
      ++cntone;
    }
  }
  return res;
}
long long getinv(vector<vector<int> >& cur, int x) {
  long long res = 0;
  for (int i = 0; i < cur.size(); ++i) {
    vector<int> nv = cur[i];
    if (x == 1) {
      for (int j = 0; j < nv.size(); ++j) {
        nv[j] = 1 - nv[j];
      }
    }
    res += getinv(nv);
  }
  return res;
}
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  long long ans = 0;
  long long val = 0;
  vector<vector<int> > cur;
  cur.push_back(a);
  for (int ps = 29; ps >= 0; --ps) {
    vector<vector<int> > nv;
    for (int i = 0; i < cur.size(); ++i) {
      vector<int> ncur(cur[i].size(), 0);
      for (int j = 0; j < cur[i].size(); ++j) {
        ncur[j] = (bool)(cur[i][j] & (1 << ps));
      }
      nv.push_back(ncur);
    }
    long long rs1 = getinv(nv, 0);
    long long rs2 = getinv(nv, 1);
    if (rs2 < rs1) {
      ans += rs2;
      val += (1 << ps);
    } else {
      ans += rs1;
    }
    vector<vector<int> > nvi;
    for (int i = 0; i < cur.size(); ++i) {
      vector<int> nv1;
      vector<int> nv2;
      for (int j = 0; j < cur[i].size(); ++j) {
        if (cur[i][j] & (1 << ps)) {
          nv1.push_back(cur[i][j]);
        } else {
          nv2.push_back(cur[i][j]);
        }
      }
      if (!nv1.empty()) {
        nvi.push_back(nv1);
      }
      if (!nv2.empty()) {
        nvi.push_back(nv2);
      }
    }
    cur = nvi;
  }
  cout << ans << " " << val << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  t = 1;
  for (int tt = 0; tt < t; ++tt) {
    solve();
  }
  return 0;
}

```