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.
mod = 1000000007
eps = 10**-9


def main():
    import sys
    input = sys.stdin.readline

    N = int(input())
    A_list = [list(map(int, input().split()))]

    cnt = 0
    x = 0
    for lv in range(30, -1, -1):
        A_list_new = []
        inv_sum = 0
        inv_max = 0
        for A in A_list:
            AA = [a >> lv & 1 for a in A]
            zero = 0
            one = 0
            inv = 0
            A_new_0 = []
            A_new_1 = []
            for i, a in enumerate(AA):
                if a == 0:
                    zero += 1
                    inv += one
                    A_new_0.append(A[i])
                else:
                    one += 1
                    A_new_1.append(A[i])
            if A_new_0:
                A_list_new.append(A_new_0)
            if A_new_1:
                A_list_new.append(A_new_1)
            inv_max += zero * one
            inv_sum += inv
        inv_flipped = inv_max - inv_sum
        if inv_sum <= inv_flipped:
            cnt += inv_sum
        else:
            cnt += inv_flipped
            x += 1 << lv
        A_list = A_list_new
    print(cnt, x)


if __name__ == '__main__':
    main()
