output_description: Print Q lines. The i-th line (1 \leq i \leq Q) should contain the answer for x
  = X_i.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N Q
      A_1 A_2 ... A_N
      X_1
      X_2
      :
      X_Q
attempt: #!/usr/bin/env python3
  
  import sys
  from typing import (
      Any,
      Callable,
      Deque,
      Dict,
      List,
      Mapping,
      Optional,
      Sequence,
      Set,
      Tuple,
      TypeVar,
      Union,
  )
  
  # import time
  # import math, cmath
  # import numpy as np
  # import scipy.sparse.csgraph as cs            # csgraph_from_dense(ndarray, null_value=inf), bellman_ford(G, return_predecessors=True), dijkstra, floyd_warshall
  # import random                                # random, uniform, randint, randrange, shuffle, sample
  # import string                                # ascii_lowercase, ascii_uppercase, ascii_letters, digits, hexdigits
  # import re                                    # re.compile(pattern) => ptn obj; p.search(s), p.match(s), p.finditer(s) => match obj; p.sub(after, s)
  from bisect import (
      bisect_left,
      bisect_right,
  )  # bisect_left(a, x, lo=0, hi=len(a)) returns i such that all(val<x for val in a[lo:i]) and all(val>-=x for val in a[i:hi]).
  
  # from collections import deque                # deque class. deque(L): dq.append(x), dq.appendleft(x), dq.pop(), dq.popleft(), dq.rotate()
  # from collections import defaultdict          # subclass of dict. defaultdict(facroty)
  # from collections import Counter              # subclass of dict. Counter(iter): c.elements(), c.most_common(n), c.subtract(iter)
  # from datetime import date, datetime          # date.today(), date(year,month,day) => date obj; datetime.now(), datetime(year,month,day,hour,second,microsecond) => datetime obj; subtraction => timedelta obj
  # from datetime.datetime import strptime       # strptime('2019/01/01 10:05:20', '%Y/%m/%d/ %H:%M:%S') returns datetime obj
  # from datetime import timedelta               # td.days, td.seconds, td.microseconds, td.total_seconds(). abs function is also available.
  # from copy import copy, deepcopy              # use deepcopy to copy multi-dimentional matrix without reference
  # from functools import reduce                 # reduce(f, iter[, init])
  # from functools import lru_cache              # @lrucache ...arguments of functions should be able to be keys of dict (e.g. list is not allowed)
  # from heapq import heapify, heappush, heappop # built-in list. heapify(L) changes list in-place to min-heap in O(n), heappush(heapL, x) and heappop(heapL) in O(lgn).
  # from heapq import _heapify_max, _heappop_max, _siftdown_max
  # from heapq import nlargest, nsmallest        # nlargest(n, iter[, key]) returns k-largest-list in O(n+klgn).
  # from itertools import count, cycle, repeat   # count(start[,step]), cycle(iter), repeat(elm[,n])
  # from itertools import groupby                # [(k, list(g)) for k, g in groupby('000112')] returns [('0',['0','0','0']), ('1',['1','1']), ('2',['2'])]
  # from itertools import starmap                # starmap(pow, [[2,5], [3,2]]) returns [32, 9]
  # from itertools import product, permutations  # product(iter, repeat=n), permutations(iter[,r])
  # from itertools import combinations, combinations_with_replacement
  # from itertools import accumulate             # accumulate(iter[, f])
  # from operator import itemgetter              # itemgetter(1), itemgetter('key')
  # from fractions import Fraction               # Fraction(a, b) => a / b ∈ Q. note: Fraction(0.1) do not returns Fraciton(1, 10). Fraction('0.1') returns Fraction(1, 10)
  
  
  def main():
      Num = Union[int, float]
      mod = 1000000007  # 10^9+7
      inf = float("inf")  # sys.float_info.max = 1.79e+308
      # inf = 2 ** 63 - 1             # (for fast JIT compile in PyPy) 9.22e+18
      sys.setrecursionlimit(10**6)  # 1000 -> 1000000
  
      def input():
          return sys.stdin.readline().rstrip()
  
      def ii():
          return int(input())
  
      def isp():
          return input().split()
  
      def mi():
          return map(int, input().split())
  
      def mi_0():
          return map(lambda x: int(x) - 1, input().split())
  
      def lmi():
          return list(map(int, input().split()))
  
      def lmi_0():
          return list(map(lambda x: int(x) - 1, input().split()))
  
      def li():
          return list(input())
  
      def debug(x):
          print(x, file=sys.stderr)
  
      # def _heappush_max(h, item): h.append(item); _siftdown_max(h, 0, len(h)-1)
  
      def calc_takahasi_top_n(x):
          # x - span <= k <= x + span なる k 軸 x 中心とした幅 span の区間を考える
          # この区間に含まれる L の要素の個数が、それよりも右側の要素の個数と一致 or 1 個少ない場合求める span である
          left = 0
          right = 10**9
          while True:
              mid = (left + right) // 2
              aoki = bisect_right(L, x + mid) - bisect_left(L, x - mid)
              takahashi = n - bisect_right(L, x + mid)
              if aoki == takahashi or aoki == takahashi - 1:
                  return takahashi
              elif aoki == takahashi + 1:
                  return takahashi + 1
              elif aoki > takahashi + 1:
                  right = mid
              else:
                  left = mid
  
      def solve(x):
          top_n = calc_takahasi_top_n(x)
          res = max(0, n - top_n * 2)
          # debug(f'top_n {top_n}')
          # debug(f'res {res}')
          return rev_acum[top_n] + skip_sum[res]
  
      n, q = mi()
      L = lmi()
      query = [ii() for _ in range(q)]
      rev_acum = [0] + list(reversed(L))
      for i in range(n):
          if i > 0:
              rev_acum[i] += rev_acum[i - 1]
      skip_sum = [0] + L[:]
      for i in range(n):
          if i > 1:
              skip_sum[i] += skip_sum[i - 2]
      # debug(rev_acum)
      # debug(skip_sum)
  
      for x in query:
          print(solve(x))
  
  
  if __name__ == "__main__":
      main()
  
problem_description: Statement
  
  There are N cards. The i-th card has an integer A_i written on it. For any two
  cards, the integers on those cards are different.
  
  Using these cards, Takahashi and Aoki will play the following game:
  
    * Aoki chooses an integer x.
    * Starting from Takahashi, the two players alternately take a card. The card should be chosen in the following manner:
      * Takahashi should take the card with the largest integer among the remaining card.
      * Aoki should take the card with the integer closest to x among the remaining card. If there are multiple such cards, he should take the card with the smallest integer among those cards.
    * The game ends when there is no card remaining.
  
  You are given Q candidates for the value of x: X_1, X_2, ..., X_Q. For each i
  (1 \leq i \leq Q), find the sum of the integers written on the cards that
  Takahashi will take if Aoki chooses x = X_i.
samples: [{"input": "5 5\n    3 5 7 11 13\n    1\n    4\n    9\n    10\n    13", "output": "31\n    31\n    27\n    23\n    23\n    \n\nFor example, when x = X_3(= 9), the game proceeds as follows:\n\n  * Takahashi takes the card with 13.\n  * Aoki takes the card with 7.\n  * Takahashi takes the card with 11.\n  * Aoki takes the card with 5.\n  * Takahashi takes the card with 3.\n\nThus, 13 + 11 + 3 = 27 should be printed on the third line.\n\n* * *"}, {"input": "4 3\n    10 20 30 40\n    2\n    34\n    34", "output": "70\n    60\n    60"}]