#include <bits/stdc++.h>
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,popcnt,abm,mmx,tune=native")
using namespace std;
template <typename T>
void uin(T &a, T b) {
  if (b < a) {
    a = b;
  }
}
template <typename T>
void uax(T &a, T b) {
  if (b > a) {
    a = b;
  }
}
const long long N = 600 * 1000 + 228;
const long long INF = 1e9 + 228;
long long n;
long long a[N], color[N];
vector<long long> g[N];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; ++i) {
    cin >> a[i];
    color[i] = 1;
  }
  long long counter = 1;
  long long first = 0;
  long long ansInv = 0;
  for (long long bit = 30; bit >= 0; --bit) {
    for (long long i = 1; i <= n; ++i) {
      g[i].clear();
    }
    for (long long i = 1; i <= n; ++i) {
      g[color[i]].emplace_back(i);
    }
    long long INV[2];
    INV[0] = INV[1] = 0;
    for (long long col = 1; col <= n; ++col) {
      if (g[col].empty()) continue;
      long long sz = g[col].size();
      long long cnt[2];
      cnt[0] = cnt[1] = 0;
      long long curInv = 0;
      long long notCurInv = 0;
      for (long long i : g[col]) {
        long long cur_bit = (a[i] >> bit) & 1;
        if (cur_bit == 0) {
          curInv += cnt[1];
        }
        if (cur_bit == 1) {
          notCurInv += cnt[0];
        }
        ++cnt[cur_bit];
      }
      INV[0] += curInv;
      INV[1] += notCurInv;
      if (cnt[0] > 0 && cnt[1] > 0) {
        ++counter;
        for (long long i : g[col]) {
          long long cur_bit = (a[i] >> bit) & 1;
          if (cur_bit == 1) {
            color[i] = counter;
          }
        }
      }
    }
    if (INV[0] > INV[1]) {
      first += (1 << bit);
    }
    ansInv += min(INV[0], INV[1]);
  }
  cout << ansInv << ' ' << first << '\n';
  return 0;
}

### 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>
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("fast-math")
#pragma GCC target("sse,sse2,sse3,ssse3,popcnt,abm,mmx,tune=native")
using namespace std;
template <typename T>
void uin(T &a, T b) {
  if (b < a) {
    a = b;
  }
}
template <typename T>
void uax(T &a, T b) {
  if (b > a) {
    a = b;
  }
}
const long long N = 600 * 1000 + 228;
const long long INF = 1e9 + 228;
long long n;
long long a[N], color[N];
vector<long long> g[N];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; ++i) {
    cin >> a[i];
    color[i] = 1;
  }
  long long counter = 1;
  long long first = 0;
  long long ansInv = 0;
  for (long long bit = 30; bit >= 0; --bit) {
    for (long long i = 1; i <= n; ++i) {
      g[i].clear();
    }
    for (long long i = 1; i <= n; ++i) {
      g[color[i]].emplace_back(i);
    }
    long long INV[2];
    INV[0] = INV[1] = 0;
    for (long long col = 1; col <= n; ++col) {
      if (g[col].empty()) continue;
      long long sz = g[col].size();
      long long cnt[2];
      cnt[0] = cnt[1] = 0;
      long long curInv = 0;
      long long notCurInv = 0;
      for (long long i : g[col]) {
        long long cur_bit = (a[i] >> bit) & 1;
        if (cur_bit == 0) {
          curInv += cnt[1];
        }
        if (cur_bit == 1) {
          notCurInv += cnt[0];
        }
        ++cnt[cur_bit];
      }
      INV[0] += curInv;
      INV[1] += notCurInv;
      if (cnt[0] > 0 && cnt[1] > 0) {
        ++counter;
        for (long long i : g[col]) {
          long long cur_bit = (a[i] >> bit) & 1;
          if (cur_bit == 1) {
            color[i] = counter;
          }
        }
      }
    }
    if (INV[0] > INV[1]) {
      first += (1 << bit);
    }
    ansInv += min(INV[0], INV[1]);
  }
  cout << ansInv << ' ' << first << '\n';
  return 0;
}

```