Dima is a beginner programmer. During his working process, he regularly has to repeat the following operation again and again: to remove every second element from the array. One day he has been bored with easy solutions of this problem, and he has come up with the following extravagant algorithm.

Let's consider that initially array contains n numbers from 1 to n and the number i is located in the cell with the index 2i - 1 (Indices are numbered starting from one) and other cells of the array are empty. Each step Dima selects a non-empty array cell with the maximum index and moves the number written in it to the nearest empty cell to the left of the selected one. The process continues until all n numbers will appear in the first n cells of the array. For example if n = 4, the array is changing as follows: [Image] 

You have to write a program that allows you to determine what number will be in the cell with index x (1 ≤ x ≤ n) after Dima's algorithm finishes.


-----Input-----

The first line contains two integers n and q (1 ≤ n ≤ 10^18, 1 ≤ q ≤ 200 000), the number of elements in the array and the number of queries for which it is needed to find the answer.

Next q lines contain integers x_{i} (1 ≤ x_{i} ≤ n), the indices of cells for which it is necessary to output their content after Dima's algorithm finishes.


-----Output-----

For each of q queries output one integer number, the value that will appear in the corresponding array cell after Dima's algorithm finishes.


-----Examples-----
Input
4 3
2
3
4

Output
3
2
4

Input
13 4
10
5
4
8

Output
13
3
8
9



-----Note-----

The first example is shown in the picture.

In the second example the final array is [1, 12, 2, 8, 3, 11, 4, 9, 5, 13, 6, 10, 7].

Difficulty: competition

Starter code:

Solutions:
import sys
[n, q] = map(int, sys.stdin.readline().strip().split())
qis = [int(sys.stdin.readline().strip()) for _ in range(q)]

def query(n, q):
	d = 2 * n - q
	while d % 2 == 0:
		d //= 2
	return (n - d // 2)

for qi in qis:
	print (query(n, qi))

import sys

def query(n, a):
    while a % 2 == 0:
        a += n - a // 2
    return a // 2 + 1

n, q = map(int, sys.stdin.readline().split())
arr = [int(sys.stdin.readline()) for _ in range(q)]
# arr = [query(n, a) for a in arr]
# sys.stdout.write(*arr)
sys.stdout.write("\n".join(str(query(n, a)) for a in arr))

import sys

def query(n, a):
    while a % 2 == 0:
        a += n - a // 2
    return a // 2 + 1

n, q = list(map(int, sys.stdin.readline().split()))
sys.stdout.write("\n".join(str(query(n, int(input()))) for _ in range(q)))
# arr = [int(sys.stdin.readline()) for _ in range(q)]
# sys.stdout.write("\n".join(str(query(n, a)) for a in arr))


import sys

def query(n, a):
    while not (a & 1):
        a += (n - a//2)
    return a+1 >> 1

n, q = map(int, sys.stdin.readline().split())
arr = [int(sys.stdin.readline()) for _ in range(q)]
sys.stdout.write("\n".join(str(query(n, a)) for a in arr))

''' CODED WITH LOVE BY SATYAM KUMAR '''

from sys import stdin, stdout
import heapq
import cProfile, math
from collections import Counter, defaultdict, deque
from bisect import bisect_left, bisect, bisect_right
import itertools
from copy import deepcopy
from fractions import Fraction
import sys, threading
import operator as op
from functools import reduce
import sys

sys.setrecursionlimit(10 ** 6)  # max depth of recursion
threading.stack_size(2 ** 27)  # new thread will get stack of such size
fac_warm_up = False
printHeap = str()
memory_constrained = False
P = 10 ** 9 + 7


class MergeFind:
    def __init__(self, n):
        self.parent = list(range(n))
        self.size = [1] * n
        self.num_sets = n
        self.lista = [[_] for _ in range(n)]

    def find(self, a):
        to_update = []
        while a != self.parent[a]:
            to_update.append(a)
            a = self.parent[a]
        for b in to_update:
            self.parent[b] = a
        return self.parent[a]

    def merge(self, a, b):
        a = self.find(a)
        b = self.find(b)
        if a == b:
            return
        if self.size[a] < self.size[b]:
            a, b = b, a
        self.num_sets -= 1
        self.parent[b] = a
        self.size[a] += self.size[b]
        self.lista[a] += self.lista[b]

    def set_size(self, a):
        return self.size[self.find(a)]

    def __len__(self):
        return self.num_sets


def display(string_to_print):
    stdout.write(str(string_to_print) + "\n")


def prime_factors(n):  # n**0.5 complex
    factors = dict()
    for i in range(2, math.ceil(math.sqrt(n)) + 1):
        while n % i == 0:
            if i in factors:
                factors[i] += 1
            else:
                factors[i] = 1
            n = n // i
    if n > 2:
        factors[n] = 1
    return (factors)


def all_factors(n):
    return set(reduce(list.__add__,
                      ([i, n // i] for i in range(1, int(n ** 0.5) + 1) if n % i == 0)))


def fibonacci_modP(n, MOD):
    if n < 2: return 1
    return (cached_fn(fibonacci_modP, (n + 1) // 2, MOD) * cached_fn(fibonacci_modP, n // 2, MOD) + cached_fn(
        fibonacci_modP, (n - 1) // 2, MOD) * cached_fn(fibonacci_modP, (n - 2) // 2, MOD)) % MOD


def factorial_modP_Wilson(n, p):
    if (p <= n):
        return 0
    res = (p - 1)
    for i in range(n + 1, p):
        res = (res * cached_fn(InverseEuler, i, p)) % p
    return res


def binary(n, digits=20):
    b = bin(n)[2:]
    b = '0' * (digits - len(b)) + b
    return b


def is_prime(n):
    """Returns True if n is prime."""
    if n < 4:
        return True
    if n % 2 == 0:
        return False
    if n % 3 == 0:
        return False
    i = 5
    w = 2
    while i * i <= n:
        if n % i == 0:
            return False
        i += w
        w = 6 - w
    return True


def generate_primes(n):
    prime = [True for i in range(n + 1)]
    p = 2
    while p * p <= n:
        if prime[p]:
            for i in range(p * 2, n + 1, p):
                prime[i] = False
        p += 1
    return prime


factorial_modP = []


def warm_up_fac(MOD):
    nonlocal factorial_modP, fac_warm_up
    if fac_warm_up: return
    factorial_modP = [1 for _ in range(fac_warm_up_size + 1)]
    for i in range(2, fac_warm_up_size):
        factorial_modP[i] = (factorial_modP[i - 1] * i) % MOD
    fac_warm_up = True


def InverseEuler(n, MOD):
    return pow(n, MOD - 2, MOD)


def nCr(n, r, MOD):
    nonlocal fac_warm_up, factorial_modP
    if not fac_warm_up:
        warm_up_fac(MOD)
        fac_warm_up = True
    return (factorial_modP[n] * (
            (pow(factorial_modP[r], MOD - 2, MOD) * pow(factorial_modP[n - r], MOD - 2, MOD)) % MOD)) % MOD


def test_print(*args):
    if testingMode:
        print(args)


def display_list(list1, sep=" "):
    stdout.write(sep.join(map(str, list1)) + "\n")


def display_2D_list(li):
    for i in li:
        print(i)


def prefix_sum(li):
    sm = 0
    res = []
    for i in li:
        sm += i
        res.append(sm)
    return res


def get_int():
    return int(stdin.readline().strip())


def get_tuple():
    return list(map(int, stdin.readline().split()))


def get_list():
    return list(map(int, stdin.readline().split()))


memory = dict()


def clear_cache():
    nonlocal memory
    memory = dict()


def cached_fn(fn, *args):
    nonlocal memory
    if args in memory:
        return memory[args]
    else:
        result = fn(*args)
        memory[args] = result
        return result


def ncr(n, r):
    return math.factorial(n) / (math.factorial(n - r) * math.factorial(r))


def binary_search(i, li):
    fn = lambda x: li[x] - x // i
    x = -1
    b = len(li)
    while b >= 1:
        while b + x < len(li) and fn(b + x) > 0:  # Change this condition 2 to whatever you like
            x += b
        b = b // 2
    return x


# -------------------------------------------------------------- MAIN PROGRAM


TestCases = False
fac_warm_up_size = 10 ** 5 + 100
optimise_for_recursion = True  # Can not be used clubbed with TestCases WHen using recursive functions, use Python 3


def main():
    n, q = get_tuple()
    diff = (n - 1) * 2 - 1
    for _ in range(q):
        num = get_int()
        if (num%2!=0):
            print((num+1)//2)
            continue
        numerator = (diff - num + 3)
        n1 = list(bin(numerator))
        k = 0
        for i in n1:
            if i=='1':
                k = 0
            else:
                k += 1
        #print(k)
        numerator = (diff-num+3)
        x = ((numerator//(2**k))-1)//2
            #print(numerator,k, diff, x)
        print(n-x)

# --------------------------------------------------------------------- END=


if TestCases:
    for i in range(get_int()):
        main()
else:
    main() if not optimise_for_recursion else threading.Thread(target=main).start()


from sys import*
input=stdin.readline
n,q=map(int,input().split())
for _ in range(q):
    i=int(input())
    while i%2==0:i+=n-i//2
    print((i+1)//2)

import sys
input=sys.stdin.readline

n,q=list(map(int,input().split()))
for i in range(q):
    a=int(input())
    if a%2!=0:
        print((a+1)//2)
    else:
        b=a//2
        while 1>0:
            a=a+(n-b)
            b=a//2
            if a%2!=0:
                print((a+1)//2)
                break