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,collections as cc
input = sys.stdin.buffer.readline
I = lambda : list(map(int,input().split()))

n,=I()
l=I()
an=ban=0
ar=cc.defaultdict(list)
ar[0]=l
xo=1<<30
for i in range(30,-1,-1):
	a=b=0
	pr=cc.defaultdict(list)
	for j in ar:
		ff=j
		j=ar[j]
		aa=bb=0
		oo=zz=0
		for x in range(len(j)):
			if j[x]&xo:
				aa+=zz
				oo+=1
				pr[2*ff+0].append(j[x])
			else:
				bb+=oo
				zz+=1
				pr[2*ff+1].append(j[x])
		a+=aa
		b+=bb
	ar=pr
	if a<b:
		an+=xo
		ban+=a
	else:
		ban+=b
	xo//=2

print(ban,an)