#include <bits/stdc++.h>
using namespace std;
const long long dx[] = {0, 1, 0, -1};
const long long dy[] = {1, 0, -1, 0};
const long long inf = 2e6;
const long long mod = 1e9 + 7;
const long long N = 3e6 + 11;
const long long INF64 = 3e18 + 1;
const double eps = 1e-14;
const double PI = acos(-1);
long long n, a[N];
pair<long long, long long> g[50];
pair<long long, long long> rec(vector<long long> v, long long bit) {
  vector<long long> b, v1, v2;
  for (long long i = 0; i < (long long)((v).size()); i++) {
    if (v[i] & (1 << bit)) {
      b.push_back(1);
      v1.push_back(v[i]);
    } else {
      b.push_back(0);
      v2.push_back(v[i]);
    }
  }
  long long x = 0, inv = 0;
  for (long long i = 0; i < (long long)((b).size()); i++) {
    if (b[i] == 0) {
      inv += x;
    } else {
      x++;
    }
  }
  long long y = 0, inv2 = 0;
  for (long long i = 0; i < (long long)((b).size()); i++) {
    if (b[i] == 1) {
      inv2 += y;
    } else {
      y++;
    }
  }
  pair<long long, long long> ans = {inv, inv2};
  if (!bit) return ans;
  pair<long long, long long> ans1 = {0, 0};
  if ((long long)((v1).size()) > 0) ans1 = rec(v1, bit - 1);
  pair<long long, long long> ans2 = {0, 0};
  if ((long long)((v2).size()) > 0) ans2 = rec(v2, bit - 1);
  g[bit - 1].first += ans1.first + ans2.first;
  g[bit - 1].second += ans1.second + ans2.second;
  return ans;
}
void solve() {
  vector<long long> base;
  cin >> n;
  long long tot = 0;
  for (long long i = 1; i <= n; ++i) {
    long long a;
    cin >> a;
    base.push_back(a);
  }
  long long rs = 0;
  auto ans = rec(base, 30);
  g[30].first += ans.first;
  g[30].second += ans.second;
  for (long long i = 0; i < 30; ++i) {
    if (g[i].first > g[i].second) {
      rs += (1ll << i);
    }
    tot += min(g[i].first, g[i].second);
  }
  cout << tot << " " << rs << "\n";
}
int32_t main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long T = 1;
  while (T--) {
    solve();
  }
}

### Prompt

Please formulate 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;
const long long dx[] = {0, 1, 0, -1};
const long long dy[] = {1, 0, -1, 0};
const long long inf = 2e6;
const long long mod = 1e9 + 7;
const long long N = 3e6 + 11;
const long long INF64 = 3e18 + 1;
const double eps = 1e-14;
const double PI = acos(-1);
long long n, a[N];
pair<long long, long long> g[50];
pair<long long, long long> rec(vector<long long> v, long long bit) {
  vector<long long> b, v1, v2;
  for (long long i = 0; i < (long long)((v).size()); i++) {
    if (v[i] & (1 << bit)) {
      b.push_back(1);
      v1.push_back(v[i]);
    } else {
      b.push_back(0);
      v2.push_back(v[i]);
    }
  }
  long long x = 0, inv = 0;
  for (long long i = 0; i < (long long)((b).size()); i++) {
    if (b[i] == 0) {
      inv += x;
    } else {
      x++;
    }
  }
  long long y = 0, inv2 = 0;
  for (long long i = 0; i < (long long)((b).size()); i++) {
    if (b[i] == 1) {
      inv2 += y;
    } else {
      y++;
    }
  }
  pair<long long, long long> ans = {inv, inv2};
  if (!bit) return ans;
  pair<long long, long long> ans1 = {0, 0};
  if ((long long)((v1).size()) > 0) ans1 = rec(v1, bit - 1);
  pair<long long, long long> ans2 = {0, 0};
  if ((long long)((v2).size()) > 0) ans2 = rec(v2, bit - 1);
  g[bit - 1].first += ans1.first + ans2.first;
  g[bit - 1].second += ans1.second + ans2.second;
  return ans;
}
void solve() {
  vector<long long> base;
  cin >> n;
  long long tot = 0;
  for (long long i = 1; i <= n; ++i) {
    long long a;
    cin >> a;
    base.push_back(a);
  }
  long long rs = 0;
  auto ans = rec(base, 30);
  g[30].first += ans.first;
  g[30].second += ans.second;
  for (long long i = 0; i < 30; ++i) {
    if (g[i].first > g[i].second) {
      rs += (1ll << i);
    }
    tot += min(g[i].first, g[i].second);
  }
  cout << tot << " " << rs << "\n";
}
int32_t main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long T = 1;
  while (T--) {
    solve();
  }
}

```