output_description: Print the minimum distance that needs to be traveled when you start at one of
  the houses and visit all the N houses.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      K N
      A_1 A_2 ... A_N
attempt: # maintain overall min
  # iterate over two loops
  # find the absolute distance and before exiting the first maintain overall min
  # also check the distance of last and first house
  def findDistBetween(i, j):
      return abs(i - j)
  
  
  def minDistfindDistBetweenLastAndFirstHouse(i, j):
      dist = findDistBetween(i, j)
      return K - dist
  
  
  K, N = input().split()
  As = input().split()
  
  K = int(K)
  N = int(N)
  
  # print ("As is")
  # print (As)
  # print (As.split())
  A = [0] * len(As)
  for i in range(len(As)):
      A[i] = int(As[i])
  
  # A = list(map(lambda x: int(x), As))
  # print (A)
  # overAllMin = float('inf')
  # minDist = 0
  # for i in range(N):
  #     minDist = 0
  #     for j in range(N):
  #         # print (i, j)
  #         minDist += findDistBetween(A[i], A[j])
  #     # find the dist b/w last and first house
  #     # minDist2 = minDistfindDistBetweenLastAndFirstHouse(A[0], A[-1])
  #     # minDist = min(minDist, minDist2)
  #     print ("Min dist", minDist)
  #     # print ("min dist2", minDist2)
  #     overAllMin = min(overAllMin, minDist)
  
  # print (overAllMin)
  overAllMin = float("inf")
  Arev = list(reversed(A))
  dist1 = 0
  dist2 = 0
  prevHouse = -1
  for i in range(N):
      # print ("PRev house", prevHouse)
      # print ("Next house", A[i])
  
      dist1 = max(findDistBetween(A[i], prevHouse), dist1)
      prevHouse = A[i]
  
  
  ma = max(dist1, minDistfindDistBetweenLastAndFirstHouse(A[0], A[-1]))
  print(K - ma)
  # for i in range(N):
  #     count = 0
  #     curHouse = A[i]
  #     j = i
  #     flag0 = 0
  #     flag1 = 0
  #     dist1 = 0
  #     dist2 = 0
  #     while count < N:
  #         if flag0 is 0:
  #             dist1 += findDistBetween(curHouse, A[j])
  #         curHouse = A[j]
  #         j += 1
  #         flag0 = 0
  #         count += 1
  
  #         if j == N and count < N:
  #             print ("in condition")
  #             j = 0
  #             dist1 += minDistfindDistBetweenLastAndFirstHouse(A[0], A[-1])
  #             flag0 = 1
  
  # count += 1
  
  # print ("dist1", dist1)
  
  # count = 0
  # curHouse = A[i]
  # j = i
  # while count < N:
  #     if flag1 is 0:
  #         dist2 += findDistBetween(curHouse, A[j])
  #     curHouse = A[j]
  #     j -= 1
  #     count += 1
  #     flag1 = 0
  #     if j == -1 and count < N:
  #         j = N - 1
  #         dist2 += minDistfindDistBetweenLastAndFirstHouse(A[0], A[-1])
  #         flag1 = 1
  #     # count += 1
  # overAllMin = min(dist1, K - findDistBetween(A[0], A[-1]), overAllMin)
  # print (dist1,  K - findDistBetween(A[0], A[-1]), overAllMin)
  
  # print (overAllMin)
  
problem_description: Statement
  
  There is a circular pond with a perimeter of K meters, and N houses around
  them.
  
  The i-th house is built at a distance of A_i meters from the northmost point
  of the pond, measured clockwise around the pond.
  
  When traveling between these houses, you can only go around the pond.
  
  Find the minimum distance that needs to be traveled when you start at one of
  the houses and visit all the N houses.
samples: [{"input": "20 3\n    5 10 15", "output": "10\n    \n\nIf you start at the 1-st house and go to the 2-nd and 3-rd houses in this\norder, the total distance traveled will be 10.\n\n* * *"}, {"input": "20 3\n    0 5 15", "output": "10\n    \n\nIf you start at the 2-nd house and go to the 1-st and 3-rd houses in this\norder, the total distance traveled will be 10."}]