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.
MB = 29

def ibit(n, k):
    return (n>>k)&1


def count(m, k):
    res0 = 0
    on = 0
    res1 = 0
    zer = 0
    prr = None
    for ell in m:
        el = ibit(ell, k)
        ot = ell >> (k+1)
        if ot != prr:
            prr = ot
            on = 0
            zer = 0
        #print(ell, el, 'ot-prr', ot, prr, 'on-zer',on, zer)
        if el == 1:
            res1 += zer
            on += 1
        else:
            res0 += on
            zer += 1
    return [res0, res1]


def diff(m, k):
    res0 = []
    res1 = []
    res = []
    prr = None
    for el in m:
        ot = el >> (k+1)
        if ot != prr:
            prr = ot
            res += res0 + res1
            res0 = []
            res1 = []
        if ibit(el, k):
            res1.append(el)
        else:
            res0.append(el)
    res += res0 + res1
    return res


n = int(input())
a = list(map(int, input().split()))

'''
f = sorted(a)
nw = f[0]
colnw = 0
NB = n*(n-1)//2
for el in f:
    if nw == el:
        colnw += 1
    else:
        NB -= colnw*(colnw-1)//2
        nw = el
        colnw = 1
'''

ll = a

bt = MB
rs = 0
inv = 0
while bt >= 0:
#    print('BT', bt, inv, rs)
    #if bt <= 3:
    #    print(bt, ll, 'b')
    ll = diff(ll, bt+1)
#    print('bt', bt, 'll', ll)
    #if bt <= 3:
    #    print(bt, ll, 'a')
    r0, r1 = count(ll, bt)
#    print('b', r0, r1, 'r0-r1')
    #if bt <= 3:
    #    print(bt, r0, r1)
    if r0 > r1:
        rs += (1 << bt)
        inv += r1
    else:
        inv += r0
#    print('t', 'r0-r1', r0, r1, 'inv-rs', inv, rs)
    bt -= 1
print(inv, rs)

