import sys
def minCoins1(arr, aim):if arr == None or len(arr) == 0 or aim < 0:return -1row = len(arr)dp = [[sys.maxsize for i in range(aim+1)] for j in range(row)]for i in range(row):dp[i][0] = 0for j in range(1, aim+1):if j % arr[0] == 0:dp[0][j] = j // arr[0]for i in range(1, row):for j in range(1, aim+1):left = sys.maxsizeif j - arr[i] >= 0 and dp[i][j-arr[i]] != sys.maxsize:left = dp[i][j-arr[i]] + 1dp[i][j] = min(left, dp[i-1][j])return dp[row-1][aim] if dp[row-1][aim] != sys.maxsize else -1
arr = [5,2,3]
aim = 20
minCoins1(arr, aim)

#经过空间压缩的动态规划
def minCoins2(arr, aim):if arr == None or len(arr) == 0 or aim < 0:return -1row = len(arr)dp = [sys.maxsize for i in range(aim+1)]dp[0] = 0for i in range(1, aim+1):if i % arr[0] == 0:dp[i] = i // arr[0]for i in range(1, row):dp[0] = 0for j in range(1, aim+1):left = sys.maxsizeif j - arr[i] >= 0 and dp[j-arr[i]] != sys.maxsize:left = dp[j-arr[i]] + 1dp[j] = min(left, dp[j])return dp[aim] if dp[aim] != sys.maxsize else -1

#暴力递归方法
def coins1(arr, aim):def process1(arr, index, aim):if index == len(arr):return 1 if aim == 0 else 0else:res = 0for i in range(0, aim//arr[index]+1):res += process1(arr, index+1, aim-arr[index]*i)return resif arr == None or len(arr) == 0 or aim < 0:return 0return process1(arr, 0, aim)

思路2：记忆搜索：时间复杂度为O(N*aim^2)，空间复杂度O(N*aim)。

采用一个字典存储计算过的结果，暴力递归会重复计算结果。比如使用0张5元+1张10元的情况和使用2张5元+0张10元的情况，都需要求[25, 1]组成剩下的990的方法数。记忆搜索就是使用一张记录表将递归过程中的结果进行记录，当下次再遇到同样的递归过程，就直接使用表中的数据。

def coins2(arr, aim):def process2(arr, index, aim, records):if index == len(arr):return 1 if aim == 0 else 0else:res = 0for i in range(0, aim//arr[index]+1):mapValue = records[index+1][aim-arr[index]*i]if mapValue != 0:res += mapValue if mapValue != -1 else 0else:res += process2(arr, index+1, aim-arr[index]*i, records)records[index][aim] = -1 if res == 0 else resreturn resif arr == None or len(arr) == 0 or aim < 0:return 0records = [[0 for i in range(aim+1)] for j in range(len(arr)+1)]return process2(arr, 0, aim, records)
arr = [5,2,3]
aim = 15
res1 = coins2(arr,aim)

def coins3(arr, aim):if arr == None or len(arr) == 0 or aim < 0:return 0row = len(arr)dp = [[0 for i in range(aim+1)]for j in range(row)]for i in range(row):dp[i][0] = 1for j in range(1, aim//arr[0]+1):dp[0][arr[0]*j] = 1for i in range(1, row):for j in range(1, aim+1):num = 0for k in range(j//arr[i]+1):num += dp[i-1][j-arr[i]*k]dp[i][j] = numreturn dp[row-1][aim]

def coins4(arr, aim):if arr == None or len(arr) == 0 or aim < 0:return 0row = len(arr)dp = [[0 for i in range(aim+1)] for j in range(row)]for i in range(row):dp[i][0] = 1for j in range(1, aim//arr[0]+1):dp[0][arr[0]*j] = 1for i in range(1,row):for j in range(1, aim+1):           dp[i][j] = dp[i-1][j]dp[i][j] += dp[i][j-arr[i]] if j-arr[i] >= 0 else 0return dp[row-1][aim]

def coin5(arr,aim):if not arr or not aim:return 0dp = [0] * (aim + 1)for i in range(0,aim+1,arr[0]):dp[i] = 1for i in range(1,len(arr)):for j in range(1,aim+1):dp[j] += dp[j-arr[i]] if j - arr[i] >= 0 else 0return dp[-1]
arr = [5,2,3]
aim = 15
coin5(arr,aim)

def test(N,M):arr = [1] * Nif not M % 2:print('No Solution')else:index = 0while 1 in arr:num1 , num2 = 0 , 0for i in range(len(arr)):if arr[i] == 1:num1 += 1num2 = N - num1print(num1,num2)end = index + M - N if (index + M - N)>0 else (index + M)if index <= end:for i in range(index,end):arr[i] = 1- arr[i]else:for i in range(0,end):arr[i] = 1-arr[i]for i in range(index,N):arr[i] = 1-arr[i]index = endprint(0,N)
N , M = input().split()
N,M = int(N),int(M)
test(N,M)