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.
n = int(input())
a = [int(x) for x in input().split()]

def solve(w, bit_level):
    l = []
    h = []

    openers = 0
    inversions = 0

    alt_openers = 0
    alt_inversions = 0
    
    for x in w:
        if x & bit_level:
            h.append(x)
            openers += 1
            alt_inversions += alt_openers
        else:
            l.append(x)
            inversions += openers
            alt_openers += 1

    return inversions, alt_inversions, l, h
        

out = 0
tot_inversions = 0
active_groups = [a]

for bit_level in reversed(range(31)):
    sum_i, sum_ai = 0, 0
    next_batch = []
    for group in active_groups:
        i, ai, l, h = solve(group, 1 << bit_level)
        sum_i += i
        sum_ai += ai
        if l: next_batch.append(l)
        if h: next_batch.append(h)

    if sum_ai < sum_i:
        out |= 1 << bit_level
        tot_inversions += sum_ai
    else:
        tot_inversions += sum_i

    active_groups = next_batch

print(tot_inversions,out)
        

    
