#include <bits/stdc++.h>
using namespace std;
vector<vector<long long>> dp(31, vector<long long>(2, 0));
void calc(vector<long long> a, long long bit) {
  if (bit == -1) return;
  vector<long long> l, r, z;
  for (long long i = 0; i < a.size(); ++i) {
    if (a[i] & (1 << bit))
      r.push_back(a[i]), z.push_back(1);
    else
      l.push_back(a[i]), z.push_back(0);
  }
  long long c1 = 0, o1 = 0;
  for (long long i = 0; i < z.size(); ++i)
    if (z[i] == 1)
      ++o1;
    else
      c1 += o1;
  long long c2 = 0, o2 = 0;
  for (long long i = 0; i < z.size(); ++i)
    if (z[i] == 0)
      ++o2;
    else
      c2 += o2;
  dp[bit][0] += c1, dp[bit][1] += c2;
  if (l.size()) calc(l, bit - 1);
  if (r.size()) calc(r, bit - 1);
  return;
}
struct FenwickTree {
  vector<long long> bit;
  long long n;
  FenwickTree(long long n) {
    this->n = n;
    bit.assign(n, 0);
  }
  FenwickTree(vector<long long> a) : FenwickTree(a.size()) {
    for (size_t i = 0; i < a.size(); i++) add(i, a[i]);
  }
  long long sum(long long r) {
    long long ret = 0;
    for (; r >= 0; r = (r & (r + 1)) - 1) ret += bit[r];
    return ret;
  }
  long long sum(long long l, long long r) { return sum(r) - sum(l - 1); }
  void add(long long idx, long long delta) {
    for (; idx < n; idx = idx | (idx + 1)) bit[idx] += delta;
  }
};
long long getInvCount(vector<long long> a) {
  map<long long, vector<long long>> m;
  for (long long i = 0; i < a.size(); ++i) m[a[i]].push_back(i);
  long long z = 0;
  for (auto it : m) {
    ++z;
    for (auto j : it.second) a[j] = z;
  }
  long long mx = *max_element(a.begin(), a.end());
  FenwickTree f(mx + 1);
  long long c = 0;
  for (long long i = a.size() - 1; i >= 0; --i) {
    c += f.sum(a[i] - 1);
    f.add(a[i], 1);
  }
  return c;
}
signed main() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; ++i) cin >> a[i];
  calc(a, 30);
  long long x = 0;
  for (long long i = 30; i >= 0; --i)
    if (dp[i][0] > dp[i][1]) x |= (1 << i);
  for (long long i = 0; i < n; ++i) a[i] = a[i] ^ x;
  long long ans = getInvCount(a);
  cout << ans << " " << x << "\n";
}

### Prompt

Develop a solution in Cpp 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<vector<long long>> dp(31, vector<long long>(2, 0));
void calc(vector<long long> a, long long bit) {
  if (bit == -1) return;
  vector<long long> l, r, z;
  for (long long i = 0; i < a.size(); ++i) {
    if (a[i] & (1 << bit))
      r.push_back(a[i]), z.push_back(1);
    else
      l.push_back(a[i]), z.push_back(0);
  }
  long long c1 = 0, o1 = 0;
  for (long long i = 0; i < z.size(); ++i)
    if (z[i] == 1)
      ++o1;
    else
      c1 += o1;
  long long c2 = 0, o2 = 0;
  for (long long i = 0; i < z.size(); ++i)
    if (z[i] == 0)
      ++o2;
    else
      c2 += o2;
  dp[bit][0] += c1, dp[bit][1] += c2;
  if (l.size()) calc(l, bit - 1);
  if (r.size()) calc(r, bit - 1);
  return;
}
struct FenwickTree {
  vector<long long> bit;
  long long n;
  FenwickTree(long long n) {
    this->n = n;
    bit.assign(n, 0);
  }
  FenwickTree(vector<long long> a) : FenwickTree(a.size()) {
    for (size_t i = 0; i < a.size(); i++) add(i, a[i]);
  }
  long long sum(long long r) {
    long long ret = 0;
    for (; r >= 0; r = (r & (r + 1)) - 1) ret += bit[r];
    return ret;
  }
  long long sum(long long l, long long r) { return sum(r) - sum(l - 1); }
  void add(long long idx, long long delta) {
    for (; idx < n; idx = idx | (idx + 1)) bit[idx] += delta;
  }
};
long long getInvCount(vector<long long> a) {
  map<long long, vector<long long>> m;
  for (long long i = 0; i < a.size(); ++i) m[a[i]].push_back(i);
  long long z = 0;
  for (auto it : m) {
    ++z;
    for (auto j : it.second) a[j] = z;
  }
  long long mx = *max_element(a.begin(), a.end());
  FenwickTree f(mx + 1);
  long long c = 0;
  for (long long i = a.size() - 1; i >= 0; --i) {
    c += f.sum(a[i] - 1);
    f.add(a[i], 1);
  }
  return c;
}
signed main() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; ++i) cin >> a[i];
  calc(a, 30);
  long long x = 0;
  for (long long i = 30; i >= 0; --i)
    if (dp[i][0] > dp[i][1]) x |= (1 << i);
  for (long long i = 0; i < n; ++i) a[i] = a[i] ^ x;
  long long ans = getInvCount(a);
  cout << ans << " " << x << "\n";
}

```