#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  return f == 1 ? x : -x;
}
const long long N = 3e5 + 4;
long long n, cur, Ans, cou, a[N], b[N], a1[N], a2[N];
vector<long long> duan[2];
inline long long solve(long long l, long long r) {
  long long ret = 0;
  for (long long i = l, t1 = 0; i <= r; i++) {
    if (b[i])
      t1++;
    else
      ret += t1;
  }
  return ret;
}
inline void split(long long l, long long r) {
  a1[0] = a2[0] = 0;
  for (long long i = l; i <= r; i++) {
    if (b[i])
      a2[++a2[0]] = a[i];
    else
      a1[++a1[0]] = a[i];
  }
  for (long long i = 1; i <= a1[0]; i++) a[i + l - 1] = a1[i];
  for (long long i = 1; i <= a2[0]; i++) a[i + l + a1[0] - 1] = a2[i];
  if (a1[0]) duan[cur].push_back(l + a1[0] - 1);
  if (a2[0]) duan[cur].push_back(r);
}
signed main() {
  n = read();
  for (long long i = 1; i <= n; i++) a[i] = read();
  duan[cur].push_back(n);
  for (long long t = 30, pre, ans0, ans1; t >= 0; t--) {
    for (long long i = 1; i <= n; i++) b[i] = (a[i] >> t) & 1;
    cur ^= 1;
    duan[cur].clear();
    pre = ans0 = 0;
    for (auto v : duan[cur ^ 1]) {
      ans0 += solve(pre + 1, v);
      pre = v;
    }
    for (long long i = 1; i <= n; i++) b[i] = b[i] ^ 1;
    pre = ans1 = 0;
    for (auto v : duan[cur ^ 1]) {
      ans1 += solve(pre + 1, v);
      pre = v;
    }
    if (ans0 <= ans1) {
      cou += ans0;
    } else {
      cou += ans1;
      Ans |= (1 << t);
    }
    pre = 0;
    for (auto v : duan[cur ^ 1]) {
      split(pre + 1, v);
      pre = v;
    }
  }
  cout << cou << " " << Ans << "\n";
  return (0 - 0);
}

### Prompt

Your task is to create 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;
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  return f == 1 ? x : -x;
}
const long long N = 3e5 + 4;
long long n, cur, Ans, cou, a[N], b[N], a1[N], a2[N];
vector<long long> duan[2];
inline long long solve(long long l, long long r) {
  long long ret = 0;
  for (long long i = l, t1 = 0; i <= r; i++) {
    if (b[i])
      t1++;
    else
      ret += t1;
  }
  return ret;
}
inline void split(long long l, long long r) {
  a1[0] = a2[0] = 0;
  for (long long i = l; i <= r; i++) {
    if (b[i])
      a2[++a2[0]] = a[i];
    else
      a1[++a1[0]] = a[i];
  }
  for (long long i = 1; i <= a1[0]; i++) a[i + l - 1] = a1[i];
  for (long long i = 1; i <= a2[0]; i++) a[i + l + a1[0] - 1] = a2[i];
  if (a1[0]) duan[cur].push_back(l + a1[0] - 1);
  if (a2[0]) duan[cur].push_back(r);
}
signed main() {
  n = read();
  for (long long i = 1; i <= n; i++) a[i] = read();
  duan[cur].push_back(n);
  for (long long t = 30, pre, ans0, ans1; t >= 0; t--) {
    for (long long i = 1; i <= n; i++) b[i] = (a[i] >> t) & 1;
    cur ^= 1;
    duan[cur].clear();
    pre = ans0 = 0;
    for (auto v : duan[cur ^ 1]) {
      ans0 += solve(pre + 1, v);
      pre = v;
    }
    for (long long i = 1; i <= n; i++) b[i] = b[i] ^ 1;
    pre = ans1 = 0;
    for (auto v : duan[cur ^ 1]) {
      ans1 += solve(pre + 1, v);
      pre = v;
    }
    if (ans0 <= ans1) {
      cou += ans0;
    } else {
      cou += ans1;
      Ans |= (1 << t);
    }
    pre = 0;
    for (auto v : duan[cur ^ 1]) {
      split(pre + 1, v);
      pre = v;
    }
  }
  cout << cou << " " << Ans << "\n";
  return (0 - 0);
}

```