output_description: Print N lines. In the i-th line, print the total cost of travel during the
  trip when the visit to Spot i is canceled.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N
      A_1 A_2 ... A_N
attempt: # -*- coding: utf-8 -*-
  
  #############
  # Libraries #
  #############
  
  import sys
  
  input = sys.stdin.readline
  
  import math
  
  # from math import gcd
  import bisect
  import heapq
  from collections import defaultdict
  from collections import deque
  from collections import Counter
  from functools import lru_cache
  
  #############
  # Constants #
  #############
  
  MOD = 10**9 + 7
  INF = float("inf")
  AZ = "abcdefghijklmnopqrstuvwxyz"
  
  #############
  # Functions #
  #############
  
  
  ######INPUT######
  def I():
      return int(input().strip())
  
  
  def S():
      return input().strip()
  
  
  def IL():
      return list(map(int, input().split()))
  
  
  def SL():
      return list(map(str, input().split()))
  
  
  def ILs(n):
      return list(int(input()) for _ in range(n))
  
  
  def SLs(n):
      return list(input().strip() for _ in range(n))
  
  
  def ILL(n):
      return [list(map(int, input().split())) for _ in range(n)]
  
  
  def SLL(n):
      return [list(map(str, input().split())) for _ in range(n)]
  
  
  ######OUTPUT######
  def P(arg):
      print(arg)
      return
  
  
  def Y():
      print("Yes")
      return
  
  
  def N():
      print("No")
      return
  
  
  def E():
      exit()
  
  
  def PE(arg):
      print(arg)
      exit()
  
  
  def YE():
      print("Yes")
      exit()
  
  
  def NE():
      print("No")
      exit()
  
  
  #####Shorten#####
  def DD(arg):
      return defaultdict(arg)
  
  
  #####Inverse#####
  def inv(n):
      return pow(n, MOD - 2, MOD)
  
  
  ######Combination######
  kaijo_memo = []
  
  
  def kaijo(n):
      if len(kaijo_memo) > n:
          return kaijo_memo[n]
      if len(kaijo_memo) == 0:
          kaijo_memo.append(1)
      while len(kaijo_memo) <= n:
          kaijo_memo.append(kaijo_memo[-1] * len(kaijo_memo) % MOD)
      return kaijo_memo[n]
  
  
  gyaku_kaijo_memo = []
  
  
  def gyaku_kaijo(n):
      if len(gyaku_kaijo_memo) > n:
          return gyaku_kaijo_memo[n]
      if len(gyaku_kaijo_memo) == 0:
          gyaku_kaijo_memo.append(1)
      while len(gyaku_kaijo_memo) <= n:
          gyaku_kaijo_memo.append(
              gyaku_kaijo_memo[-1] * pow(len(gyaku_kaijo_memo), MOD - 2, MOD) % MOD
          )
      return gyaku_kaijo_memo[n]
  
  
  def nCr(n, r):
      if n == r:
          return 1
      if n < r or r < 0:
          return 0
      ret = 1
      ret = ret * kaijo(n) % MOD
      ret = ret * gyaku_kaijo(r) % MOD
      ret = ret * gyaku_kaijo(n - r) % MOD
      return ret
  
  
  ######Factorization######
  def factorization(n):
      arr = []
      temp = n
      for i in range(2, int(-(-(n**0.5) // 1)) + 1):
          if temp % i == 0:
              cnt = 0
              while temp % i == 0:
                  cnt += 1
                  temp //= i
              arr.append([i, cnt])
      if temp != 1:
          arr.append([temp, 1])
      if arr == []:
          arr.append([n, 1])
      return arr
  
  
  #####MakeDivisors######
  def make_divisors(n):
      divisors = []
      for i in range(1, int(n**0.5) + 1):
          if n % i == 0:
              divisors.append(i)
              if i != n // i:
                  divisors.append(n // i)
      return divisors
  
  
  #####MakePrimes######
  def make_primes(N):
      max = int(math.sqrt(N))
      seachList = [i for i in range(2, N + 1)]
      primeNum = []
      while seachList[0] <= max:
          primeNum.append(seachList[0])
          tmp = seachList[0]
          seachList = [i for i in seachList if i % tmp != 0]
      primeNum.extend(seachList)
      return primeNum
  
  
  #####GCD#####
  def gcd(a, b):
      while b:
          a, b = b, a % b
      return a
  
  
  #####LCM#####
  def lcm(a, b):
      return a * b // gcd(a, b)
  
  
  #####BitCount#####
  def count_bit(n):
      count = 0
      while n:
          n &= n - 1
          count += 1
      return count
  
  
  #####ChangeBase#####
  def base_10_to_n(X, n):
      if X // n:
          return base_10_to_n(X // n, n) + [X % n]
      return [X % n]
  
  
  def base_n_to_10(X, n):
      return sum(int(str(X)[-i - 1]) * n**i for i in range(len(str(X))))
  
  
  def base_10_to_n_without_0(X, n):
      X -= 1
      if X // n:
          return base_10_to_n_without_0(X // n, n) + [X % n]
      return [X % n]
  
  
  #####IntLog#####
  def int_log(n, a):
      count = 0
      while n >= a:
          n //= a
          count += 1
      return count
  
  
  #############
  # Main Code #
  #############
  N = I()
  A = IL()
  
  POS = 0
  SUM = 0
  for a in A:
      SUM += abs(POS - a)
      POS = a
  SUM += abs(POS)
  
  for i in range(N):
      if i != 0 and i != N - 1:
          print(
              SUM - abs(A[i] - A[i - 1]) - abs(A[i] - A[i + 1]) + abs(A[i - 1] - A[i + 1])
          )
      if i == 0:
          print(SUM - abs(A[i]) - abs(A[i] - A[i + 1]) + abs(A[i + 1]))
      if i == N - 1:
          print(SUM - abs(A[i] - A[i - 1]) - abs(A[i]) + abs(A[i - 1]))
  
problem_description: Statement
  
  There are N sightseeing spots on the x-axis, numbered 1, 2, ..., N. Spot i is
  at the point with coordinate A_i. It costs |a - b| yen (the currency of Japan)
  to travel from a point with coordinate a to another point with coordinate b
  along the axis.
  
  You planned a trip along the axis. In this plan, you first depart from the
  point with coordinate 0, then visit the N spots in the order they are
  numbered, and finally return to the point with coordinate 0.
  
  However, something came up just before the trip, and you no longer have enough
  time to visit all the N spots, so you decided to choose some i and cancel the
  visit to Spot i. You will visit the remaining spots as planned in the order
  they are numbered. You will also depart from and return to the point with
  coordinate 0 at the beginning and the end, as planned.
  
  For each i = 1, 2, ..., N, find the total cost of travel during the trip when
  the visit to Spot i is canceled.
samples: [{"input": "3\n    3 5 -1", "output": "12\n    8\n    10\n    \n\nSpot 1, 2 and 3 are at the points with coordinates 3, 5 and -1, respectively.\nFor each i, the course of the trip and the total cost of travel when the visit\nto Spot i is canceled, are as follows:\n\n  * For i = 1, the course of the trip is 0 \\rightarrow 5 \\rightarrow -1 \\rightarrow 0 and the total cost of travel is 5 + 6 + 1 = 12 yen.\n  * For i = 2, the course of the trip is 0 \\rightarrow 3 \\rightarrow -1 \\rightarrow 0 and the total cost of travel is 3 + 4 + 1 = 8 yen.\n  * For i = 3, the course of the trip is 0 \\rightarrow 3 \\rightarrow 5 \\rightarrow 0 and the total cost of travel is 3 + 2 + 5 = 10 yen.\n\n* * *"}, {"input": "5\n    1 1 1 2 0", "output": "4\n    4\n    4\n    2\n    4\n    \n\n* * *"}, {"input": "6\n    -679 -2409 -3258 3095 -3291 -4462", "output": "21630\n    21630\n    19932\n    8924\n    21630\n    19288"}]