#include <bits/stdc++.h>
using namespace std;
void setIO(string str, bool dbg) {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(nullptr);
  if (!dbg) {
    freopen((str + ".in").c_str(), "r", stdin);
    freopen((str + ".out").c_str(), "w", stdout);
  }
}
const int MAX = 6e6 + 5;
const int LEN = 3e1 + 5;
const int LVL = 2e1 + 0;
const long long int MOD = 1e9 + 7;
const long long int INF = 2e9;
int xyz = 1;
int lbl = 1;
long long int dps[LEN][2];
int con[MAX][2];
vector<int> sub[MAX];
void add(int num, int idx) {
  int cur = 0;
  for (int i = 30; i >= 0; i--) {
    int bit = (num >> i) & 1;
    if (con[cur][bit] == 0) con[cur][bit] = lbl++;
    sub[cur = con[cur][bit]].push_back(idx);
  }
}
void dfs(int pos, int sft) {
  int l = con[pos][0];
  int r = con[pos][1];
  if (l) dfs(l, sft - 1);
  if (r) dfs(r, sft - 1);
  if (!l || !r) return;
  long long int res = 0;
  int lpt = 0;
  int rpt = 0;
  for (; lpt < sub[l].size(); lpt++) {
    while (rpt < sub[r].size() && sub[r][rpt] < sub[l][lpt]) rpt++;
    res += rpt;
  }
  dps[sft][0] += res;
  dps[sft][1] += 1ll * sub[l].size() * sub[r].size() - res;
}
void fnc() {
  int n;
  cin >> n;
  int inp;
  for (int i = 1; i <= n; i++) {
    cin >> inp;
    add(inp, i);
  }
  dfs(0, 30);
  long long int inv = 0;
  int res = 0;
  for (int i = 0; i <= 30; i++) {
    inv += min(dps[i][0], dps[i][1]);
    if (dps[i][1] < dps[i][0]) res |= (1 << i);
  }
  cout << inv << " " << res << "\n";
}
int main() {
  setIO("", 1);
  while (xyz--) fnc();
  return 0;
}

### Prompt

Generate 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;
void setIO(string str, bool dbg) {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(nullptr);
  if (!dbg) {
    freopen((str + ".in").c_str(), "r", stdin);
    freopen((str + ".out").c_str(), "w", stdout);
  }
}
const int MAX = 6e6 + 5;
const int LEN = 3e1 + 5;
const int LVL = 2e1 + 0;
const long long int MOD = 1e9 + 7;
const long long int INF = 2e9;
int xyz = 1;
int lbl = 1;
long long int dps[LEN][2];
int con[MAX][2];
vector<int> sub[MAX];
void add(int num, int idx) {
  int cur = 0;
  for (int i = 30; i >= 0; i--) {
    int bit = (num >> i) & 1;
    if (con[cur][bit] == 0) con[cur][bit] = lbl++;
    sub[cur = con[cur][bit]].push_back(idx);
  }
}
void dfs(int pos, int sft) {
  int l = con[pos][0];
  int r = con[pos][1];
  if (l) dfs(l, sft - 1);
  if (r) dfs(r, sft - 1);
  if (!l || !r) return;
  long long int res = 0;
  int lpt = 0;
  int rpt = 0;
  for (; lpt < sub[l].size(); lpt++) {
    while (rpt < sub[r].size() && sub[r][rpt] < sub[l][lpt]) rpt++;
    res += rpt;
  }
  dps[sft][0] += res;
  dps[sft][1] += 1ll * sub[l].size() * sub[r].size() - res;
}
void fnc() {
  int n;
  cin >> n;
  int inp;
  for (int i = 1; i <= n; i++) {
    cin >> inp;
    add(inp, i);
  }
  dfs(0, 30);
  long long int inv = 0;
  int res = 0;
  for (int i = 0; i <= 30; i++) {
    inv += min(dps[i][0], dps[i][1]);
    if (dps[i][1] < dps[i][0]) res |= (1 << i);
  }
  cout << inv << " " << res << "\n";
}
int main() {
  setIO("", 1);
  while (xyz--) fnc();
  return 0;
}

```