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

que = [A, 1]

ans = 0
ans1 = 0

p = 29

for p in range(29, -1, -1):
    X1 = 0
    Y1 = 0
    
    r = 2 ** p
    q1 = que.pop()
    q2 = 0
    que1 = []
    
    for i in range(q1):
        x = 0
        y = 0
        X = 0
        Y = 0
        A1 = []
        A2 = []
        a = que.pop()
        for j in a:
            if j < r:
                X += x
                y += 1
                A1 += [j]
            else:
                x += 1
                Y += y
                A2 += [j - r]
        if A1:
            q2 += 1
            que1 += [A1]
        if A2:
            q2 += 1
            que1 += [A2]
        X1 += X
        Y1 += Y
    if X1 > Y1:
        ans += r
    ans1 += min(X1, Y1)
    que += que1
    que += [q2]
print(ans1, ans)