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.readline
N, = map(int, input().split())
A = list(map(int, input().split()))

def check(l, i):
    al,bl=[],[]
    a=b=0
    ta=tb=0
    for j in l:
        if (A[j]>>i) & 1:
            al.append(j)
            ta += 1
            b += tb
        else:
            bl.append(j)
            tb += 1
            a += ta
    return a,b,al, bl


ll = [list(range(N))]
R = 0
X = 0
for i in range(29, -1, -1):
    nll = []
    ac=bc=0
#    print(i, ll)
    for l in ll:
        a, b, al, bl = check(l, i)
        ac += a
        bc += b
        if al:
            nll.append(al)
        if bl:
            nll.append(bl)
#    print(ac, bc)
    if ac > bc:
        X += 1<<i
        R += bc
    else:
        R += ac
    ll = nll
print(R, X)
