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 io
import os

from collections import Counter, defaultdict, deque


def solve(N, A):
    total = 0
    ans = 0
    groups = [A]  # group by high bits equal so far
    for pos in reversed(range(32)):
        bit = 1 << pos

        invs = 0
        flippedInvs = 0
        nextGroups = []
        for group in groups:
            binary = []
            big = []
            small = []
            for x in group:
                if x & bit:
                    binary.append(1)
                    big.append(x)
                else:
                    binary.append(0)
                    small.append(x)
            if big:
                nextGroups.append(big)
            if small:
                nextGroups.append(small)

            numZeroes = 0
            numOnes = 0
            for x in binary[::-1]:
                if x == 0:
                    numZeroes += 1
                    flippedInvs += numOnes
                else:
                    numOnes += 1
                    invs += numZeroes
        groups = nextGroups
        if flippedInvs < invs:
            ans |= 1 << pos
        total += min(flippedInvs, invs)

    return str(total) + " " + str(ans)


if __name__ == "__main__":
    input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

    (N,) = [int(x) for x in input().split()]
    A = [int(x) for x in input().split()]
    ans = solve(N, A)
    print(ans)
