#include <bits/stdc++.h>
using namespace std;
long long p1 = 1000000, p2 = 0;
char buf[1000005], wb[1000005];
long long getint() {
  long long ret = 0, flag = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') flag = -1;
    c = getchar();
  }
  while (c <= '9' && c >= '0') {
    ret = (ret << 3) + (ret << 1) + c - '0';
    c = getchar();
  }
  return ret * flag;
}
void pc(char x) {
  if (p2 >= 1000000) fwrite(wb, 1, 1000000, stdout), p2 = 0;
  wb[p2++] = x;
}
void wrt(long long x) {
  if (x < 0) pc('-'), x = -x;
  long long c[24] = {0};
  if (!x) return pc('0'), void();
  while (x) c[++c[0]] = x % 10, x /= 10;
  while (c[0]) pc(c[c[0]--] + '0');
}
long long n, a[300005], sa[33], sd[33], ans, sum;
void CDQ(long long l, long long r, long long dep) {
  if (l >= r || dep < 0) return;
  long long r1 = 0, s1 = 0, r2 = 0, s2 = 0;
  for (long long i = l; i <= r; i++) {
    if ((a[i] >> dep) & 1)
      r2 = r2 + s2, ++s1;
    else
      r1 = r1 + s1, ++s2;
  }
  sa[dep] += r1;
  sd[dep] += r2;
  static long long tmpl[300005] = {0}, tmpr[300005];
  tmpl[0] = tmpr[0] = 0;
  for (long long i = l; i <= r; i++) {
    if ((a[i] >> dep) & 1)
      tmpr[++tmpr[0]] = a[i];
    else
      tmpl[++tmpl[0]] = a[i];
  }
  long long j = l;
  for (long long i = 1; i <= tmpl[0]; i++) a[j++] = tmpl[i];
  for (long long i = 1; i <= tmpr[0]; i++) a[j++] = tmpr[i];
  long long mid = l + tmpl[0] - 1;
  CDQ(l, mid, dep - 1);
  CDQ(mid + 1, r, dep - 1);
}
int main() {
  n = getint();
  for (long long i = 1; i <= n; i++) a[i] = getint();
  CDQ(1, n, 30);
  for (long long i = 30; i >= 0; i--) {
    if (sa[i] > sd[i])
      ans |= 1 << i, sum += sd[i];
    else
      sum += sa[i];
  }
  wrt(sum), pc(' ');
  wrt(ans), pc('\n');
  fwrite(wb, 1, p2, stdout);
  return 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;
long long p1 = 1000000, p2 = 0;
char buf[1000005], wb[1000005];
long long getint() {
  long long ret = 0, flag = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') flag = -1;
    c = getchar();
  }
  while (c <= '9' && c >= '0') {
    ret = (ret << 3) + (ret << 1) + c - '0';
    c = getchar();
  }
  return ret * flag;
}
void pc(char x) {
  if (p2 >= 1000000) fwrite(wb, 1, 1000000, stdout), p2 = 0;
  wb[p2++] = x;
}
void wrt(long long x) {
  if (x < 0) pc('-'), x = -x;
  long long c[24] = {0};
  if (!x) return pc('0'), void();
  while (x) c[++c[0]] = x % 10, x /= 10;
  while (c[0]) pc(c[c[0]--] + '0');
}
long long n, a[300005], sa[33], sd[33], ans, sum;
void CDQ(long long l, long long r, long long dep) {
  if (l >= r || dep < 0) return;
  long long r1 = 0, s1 = 0, r2 = 0, s2 = 0;
  for (long long i = l; i <= r; i++) {
    if ((a[i] >> dep) & 1)
      r2 = r2 + s2, ++s1;
    else
      r1 = r1 + s1, ++s2;
  }
  sa[dep] += r1;
  sd[dep] += r2;
  static long long tmpl[300005] = {0}, tmpr[300005];
  tmpl[0] = tmpr[0] = 0;
  for (long long i = l; i <= r; i++) {
    if ((a[i] >> dep) & 1)
      tmpr[++tmpr[0]] = a[i];
    else
      tmpl[++tmpl[0]] = a[i];
  }
  long long j = l;
  for (long long i = 1; i <= tmpl[0]; i++) a[j++] = tmpl[i];
  for (long long i = 1; i <= tmpr[0]; i++) a[j++] = tmpr[i];
  long long mid = l + tmpl[0] - 1;
  CDQ(l, mid, dep - 1);
  CDQ(mid + 1, r, dep - 1);
}
int main() {
  n = getint();
  for (long long i = 1; i <= n; i++) a[i] = getint();
  CDQ(1, n, 30);
  for (long long i = 30; i >= 0; i--) {
    if (sa[i] > sd[i])
      ans |= 1 << i, sum += sd[i];
    else
      sum += sa[i];
  }
  wrt(sum), pc(' ');
  wrt(ans), pc('\n');
  fwrite(wb, 1, p2, stdout);
  return 0;
}

```