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.
import sys
input = sys.stdin.buffer.readline

n, ls = int(input()), list(map(int, input().split()))
cur, res, cc = [ls], 0, 0
for i in range(30, -1, -1):
    t_rc0, t_rc1, nx = 0, 0, []
    for ar in cur:
        ar0, ar1 = [], []
        rc0, cc0, rc1, cc1 = 0, 0, 0, 0
        for u in ar:
            b = u & (1<<i)
            if b:
                rc1, cc1 = rc1+cc0, cc1+1
                ar1.append(u)
            else:
                rc0, cc0 = rc0+cc1, cc0+1
                ar0.append(u)
        t_rc0, t_rc1 = t_rc0+rc0, t_rc1+rc1
        if ar0: nx.append(ar0)
        if ar1: nx.append(ar1)
    if t_rc0 > t_rc1: res |= (1<<i)
    cc += min(t_rc0, t_rc1)
    cur = nx

print(cc, res)
