#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const int P = 1000000007;
const int V = 300100;
const int M = 30;
int p[M + 10];
int a[V];
vector<int> g[2][V], b[2];
int cnt[2];
int n;
int main() {
  p[0] = 1;
  for (int i = 1; i <= M; ++i) p[i] = p[i - 1] * 2;
  while (~scanf("%d", &n)) {
    for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    long long inv = 0;
    int ans = 0;
    int cur = 0, nxt = -1;
    g[0][0].clear();
    for (int i = 1; i <= n; ++i) g[0][0].push_back(a[i]);
    cnt[cur] = 1;
    for (int i = M; i >= 0; --i) {
      nxt = 1 - cur;
      cnt[nxt] = 0;
      long long old = 0;
      long long total = 0;
      for (int j = 0; j < cnt[cur]; ++j) {
        int c1 = 0;
        b[0].clear(), b[1].clear();
        for (int k = 0; k < g[cur][j].size(); ++k) {
          if (g[cur][j][k] & (1 << i))
            ++c1, b[1].push_back(g[cur][j][k]);
          else
            old += c1, b[0].push_back(g[cur][j][k]);
        }
        total += (long long)b[0].size() * b[1].size();
        for (int k = 0; k < 2; ++k) {
          if (b[k].size() == 0) continue;
          g[nxt][cnt[nxt]].clear();
          for (int l = 0; l < b[k].size(); ++l)
            g[nxt][cnt[nxt]].push_back(b[k][l]);
          ++cnt[nxt];
        }
      }
      if (old * 2 > total)
        ans += p[i], inv += total - old;
      else
        inv += old;
      cur = nxt;
    }
    printf("%I64d %d\n", inv, ans);
  }
  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>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const int P = 1000000007;
const int V = 300100;
const int M = 30;
int p[M + 10];
int a[V];
vector<int> g[2][V], b[2];
int cnt[2];
int n;
int main() {
  p[0] = 1;
  for (int i = 1; i <= M; ++i) p[i] = p[i - 1] * 2;
  while (~scanf("%d", &n)) {
    for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    long long inv = 0;
    int ans = 0;
    int cur = 0, nxt = -1;
    g[0][0].clear();
    for (int i = 1; i <= n; ++i) g[0][0].push_back(a[i]);
    cnt[cur] = 1;
    for (int i = M; i >= 0; --i) {
      nxt = 1 - cur;
      cnt[nxt] = 0;
      long long old = 0;
      long long total = 0;
      for (int j = 0; j < cnt[cur]; ++j) {
        int c1 = 0;
        b[0].clear(), b[1].clear();
        for (int k = 0; k < g[cur][j].size(); ++k) {
          if (g[cur][j][k] & (1 << i))
            ++c1, b[1].push_back(g[cur][j][k]);
          else
            old += c1, b[0].push_back(g[cur][j][k]);
        }
        total += (long long)b[0].size() * b[1].size();
        for (int k = 0; k < 2; ++k) {
          if (b[k].size() == 0) continue;
          g[nxt][cnt[nxt]].clear();
          for (int l = 0; l < b[k].size(); ++l)
            g[nxt][cnt[nxt]].push_back(b[k][l]);
          ++cnt[nxt];
        }
      }
      if (old * 2 > total)
        ans += p[i], inv += total - old;
      else
        inv += old;
      cur = nxt;
    }
    printf("%I64d %d\n", inv, ans);
  }
  return 0;
}

```