#include <bits/stdc++.h>
using namespace std;
using ll = long long int;
ll date[1000000 + 110];
ll Msort(ll l, ll r) {
  static ll c[1000000 + 110];
  ll ret = 0;
  ll mid = (l + r) / 2;
  ll i, j, tmp;
  if (r > l) {
    ret += Msort(l, mid);
    ret += Msort(mid + 1, r);
    tmp = l;
    for (i = l, j = mid + 1; i <= mid && j <= r;) {
      if (date[i] > date[j]) {
        c[tmp++] = date[j++];
        ret += mid - i + 1;
      } else {
        c[tmp++] = date[i++];
      }
    }
    if (i <= mid)
      for (; i <= mid;) c[tmp++] = date[i++];
    if (j <= r)
      for (; j <= r;) c[tmp++] = date[j++];
    for (i = l; i <= r; i++) date[i] = c[i];
  }
  return ret;
}
ll merge(ll star, ll fini, ll val) {
  static ll t[300000 + 110];
  for (ll i = star; i <= fini; i++) t[i] = date[i], date[i] ^= val;
  ll ret = Msort(star, fini);
  for (ll i = star; i <= fini; i++) date[i] = t[i];
  return ret;
}
int main() {
  ll n;
  scanf("%lld", &n);
  for (ll i = 1; i <= n; i++) scanf("%lld", date + i);
  ll maxv = *max_element(date + 1, date + 1 + n);
  ll bit = 1;
  while ((1ll << bit) <= maxv) bit++;
  ll ret = 0;
  ll lasta = 0, lastb = 0;
  ll lastv = merge(1, n, 0);
  for (ll i = bit; i >= 0; i--) {
    ll a, b;
    a = lastv;
    b = merge(1, n, ret ^ (1ll << i));
    if (a > b) lastv = b, ret ^= (1ll << i);
  }
  cout << merge(1, n, ret) << " " << ret;
  return 0;
}

### 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;
using ll = long long int;
ll date[1000000 + 110];
ll Msort(ll l, ll r) {
  static ll c[1000000 + 110];
  ll ret = 0;
  ll mid = (l + r) / 2;
  ll i, j, tmp;
  if (r > l) {
    ret += Msort(l, mid);
    ret += Msort(mid + 1, r);
    tmp = l;
    for (i = l, j = mid + 1; i <= mid && j <= r;) {
      if (date[i] > date[j]) {
        c[tmp++] = date[j++];
        ret += mid - i + 1;
      } else {
        c[tmp++] = date[i++];
      }
    }
    if (i <= mid)
      for (; i <= mid;) c[tmp++] = date[i++];
    if (j <= r)
      for (; j <= r;) c[tmp++] = date[j++];
    for (i = l; i <= r; i++) date[i] = c[i];
  }
  return ret;
}
ll merge(ll star, ll fini, ll val) {
  static ll t[300000 + 110];
  for (ll i = star; i <= fini; i++) t[i] = date[i], date[i] ^= val;
  ll ret = Msort(star, fini);
  for (ll i = star; i <= fini; i++) date[i] = t[i];
  return ret;
}
int main() {
  ll n;
  scanf("%lld", &n);
  for (ll i = 1; i <= n; i++) scanf("%lld", date + i);
  ll maxv = *max_element(date + 1, date + 1 + n);
  ll bit = 1;
  while ((1ll << bit) <= maxv) bit++;
  ll ret = 0;
  ll lasta = 0, lastb = 0;
  ll lastv = merge(1, n, 0);
  for (ll i = bit; i >= 0; i--) {
    ll a, b;
    a = lastv;
    b = merge(1, n, ret ^ (1ll << i));
    if (a > b) lastv = b, ret ^= (1ll << i);
  }
  cout << merge(1, n, ret) << " " << ret;
  return 0;
}

```