1417_E. XOR Inverse
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.
```python
import os
from sys import stdin, stdout

class Input:
    def __init__(self):
        self.lines = stdin.readlines()
        self.idx = 0
    
    def line(self):
        try:
            return self.lines[self.idx].strip()
        finally:
            self.idx += 1
    
    def array(self, sep = ' ', cast = int):
        return list(map(cast, self.line().split(sep = sep)))
    
    def known_tests(self):
        num_of_cases, = self.array()
    
        for case in range(num_of_cases):
            yield self
    
    def unknown_tests(self):
        while self.idx < len(self.lines):
            yield self

def problem_solver():
    oo = float('Inf')
    mem = None

    def backtrack(a, L, R, B = 31):
        if B < 0 or L == R:
            return

        mask = 1 << B
        l = L
        r = L
        inv0 = 0
        inv1 = 0
        zero = 0
        one = 0
        la = []
        ra = []
        while l < R:
            while r < R and (a[l] & mask) == (a[r] & mask):
                if (a[r] & mask):
                    la.append(a[r])
                    one += 1
                    inv1 += zero
                else:
                    ra.append(a[r])
                    zero += 1
                    inv0 += one

                r += 1

            l = r
        
        backtrack(la, 0, len(la), B - 1)
        backtrack(ra, 0, len(ra), B - 1)

        mem[B][0] += inv0
        mem[B][1] += inv1

    '''
    
    '''
    def solver(inpt):
        nonlocal mem

        n, = inpt.array()
        a = inpt.array()

        mem = [[0, 0] for i in range(32)]

        backtrack(a, 0, n)

        inv = 0
        x = 0

        for b in range(32):
            if mem[b][0] <= mem[b][1]:
                inv += mem[b][0]
            else:
                inv += mem[b][1]
                x |= 1 << b

        print(inv, x)

    '''Returns solver'''
    return solver

try:
    solver = problem_solver()
    for tc in Input().unknown_tests():
        solver(tc)
except Exception as e:
    import traceback
    traceback.print_exc(file=stdout)
```