#k桁目まで見たとき、1の個数がiこ、今j個1が連続しているものの個数
N,K=map(int,input().split())
s=bin(N)[2:]
t=bin(K)[2:]
mod=998244353
if len(s)<len(t):
  s=(len(t)-len(s))*"0"+s
else:
  t=(len(s)-len(t))*"0"+t
M=len(s)
dp=[[[0]*(M+1) for i in range(M+1)] for j in range(M+1)]
dp[0][0][0]=1
for k in range(M):
  for i in range(M+1):
    for j in range(M+1):
      if t[k]=="1":
        a=1
      else:
        a=0
      if s[k]=="0" and t[k]=="0":
        dp[k+1][i][0]+=dp[k][i][j]
      elif s[k]=="0" and t[k]=="1":
        if i+1<=M and j+1<=M:
          dp[k+1][i+1][j+1]+=dp[k][i][j]
      elif s[k]=="1" and t[k]=="0":
        dp[k+1][i][0]+=dp[k][i][j]
        if i+1<=M and j+1<=M:
          dp[k+1][i+1][j+1]+=dp[k][i][j]
      else:
        if i+1<=M and j+1<=M:
          dp[k+1][i+1][j+1]+=dp[k][i][j]
        if 0<=i-j+1<=M:
          dp[k+1][i-j+1][0]+=dp[k][i][j]
ans=0
for i in range(M+1):
  for j in range(M+1):
    ans+=pow(2,i,mod)*dp[M][i][j]
    ans%=mod
print(ans)