#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 7;
const int MAXN = 8e6 + 7;
const int INF = 1e9 + 7;
const int MOD = 1e9 + 7;
int a[maxn];
int b[maxn];
int c[maxn];
int d[MAXN];
long long tmp = 0;
void merges(int l, int mid, int r) {
  int i = l, j = mid + 1;
  int now = l;
  while (i <= mid && j <= r) {
    if (b[i] <= b[j]) {
      c[now++] = b[i++];
    } else {
      c[now++] = b[j++];
      tmp += mid - i + 1;
    }
  }
  while (i <= mid) {
    c[now++] = b[i++];
  }
  while (j <= r) {
    c[now++] = b[j++];
  }
  for (i = l; i <= r; i++) b[i] = c[i];
}
void merge(int l, int r) {
  if (l == r) return;
  int mid = (l + r) >> 1;
  merge(l, mid);
  merge(mid + 1, r);
  merges(l, mid, r);
  return;
}
map<int, int> mp;
int main() {
  int i, j, n, m, t, z;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[i] = a[i];
  }
  merge(1, n);
  int now = 0;
  int ans = 0;
  for (i = 0; i <= 29; i++) {
    mp.clear();
    long long tmpp = 0;
    if (i <= 7) {
      for (j = 1; j <= n; j++) {
        mp[a[j]]++;
        int cnt = mp[a[j] ^ 1];
        if (cnt) {
          if (a[j] & 1) {
            tmpp += cnt;
          } else {
            tmpp -= cnt;
          }
        }
        a[j] = a[j] >> 1;
      }
    } else {
      for (j = 1; j <= n; j++) {
        d[a[j]]++;
        int cnt = d[a[j] ^ 1];
        if (cnt) {
          if (a[j] & 1) {
            tmpp += cnt;
          } else {
            tmpp -= cnt;
          }
        }
      }
      for (j = 1; j <= n; j++) {
        d[a[j]]--;
        a[j] = a[j] >> 1;
      }
    }
    if (tmpp < 0) {
      tmp += tmpp;
      ans += (1 << i);
    }
  }
  printf("%lld %d\n", tmp, 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>
using namespace std;
const int maxn = 3e5 + 7;
const int MAXN = 8e6 + 7;
const int INF = 1e9 + 7;
const int MOD = 1e9 + 7;
int a[maxn];
int b[maxn];
int c[maxn];
int d[MAXN];
long long tmp = 0;
void merges(int l, int mid, int r) {
  int i = l, j = mid + 1;
  int now = l;
  while (i <= mid && j <= r) {
    if (b[i] <= b[j]) {
      c[now++] = b[i++];
    } else {
      c[now++] = b[j++];
      tmp += mid - i + 1;
    }
  }
  while (i <= mid) {
    c[now++] = b[i++];
  }
  while (j <= r) {
    c[now++] = b[j++];
  }
  for (i = l; i <= r; i++) b[i] = c[i];
}
void merge(int l, int r) {
  if (l == r) return;
  int mid = (l + r) >> 1;
  merge(l, mid);
  merge(mid + 1, r);
  merges(l, mid, r);
  return;
}
map<int, int> mp;
int main() {
  int i, j, n, m, t, z;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[i] = a[i];
  }
  merge(1, n);
  int now = 0;
  int ans = 0;
  for (i = 0; i <= 29; i++) {
    mp.clear();
    long long tmpp = 0;
    if (i <= 7) {
      for (j = 1; j <= n; j++) {
        mp[a[j]]++;
        int cnt = mp[a[j] ^ 1];
        if (cnt) {
          if (a[j] & 1) {
            tmpp += cnt;
          } else {
            tmpp -= cnt;
          }
        }
        a[j] = a[j] >> 1;
      }
    } else {
      for (j = 1; j <= n; j++) {
        d[a[j]]++;
        int cnt = d[a[j] ^ 1];
        if (cnt) {
          if (a[j] & 1) {
            tmpp += cnt;
          } else {
            tmpp -= cnt;
          }
        }
      }
      for (j = 1; j <= n; j++) {
        d[a[j]]--;
        a[j] = a[j] >> 1;
      }
    }
    if (tmpp < 0) {
      tmp += tmpp;
      ans += (1 << i);
    }
  }
  printf("%lld %d\n", tmp, ans);
  return 0;
}

```