You are given two positive integers N and K. How many multisets of rational numbers satisfy all of the following conditions?
 - The multiset has exactly N elements and the sum of them is equal to K.
 - Each element of the multiset is one of 1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \dots. In other words, each element can be represented as \frac{1}{2^i}\ (i = 0,1,\dots).
The answer may be large, so print it modulo 998244353.

-----Constraints-----
 - 1 \leq K \leq N \leq 3000
 - All values in input are integers.

-----Input-----
Input is given from Standard Input in the following format:
N K

-----Output-----
Print the number of multisets of rational numbers that satisfy all of the given conditions modulo 998244353.

-----Sample Input-----
4 2

-----Sample Output-----
2

The following two multisets satisfy all of the given conditions:
 - {1, \frac{1}{2}, \frac{1}{4}, \frac{1}{4}}
 - {\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}}

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = 'import sys\ninput = sys.stdin.readline\n\nsys.setrecursionlimit(10000)\n\ndef I(): return int(input())\ndef MI(): return list(map(int, input().split()))\ndef LI(): return list(map(int, input().split()))\nmod=998244353\n\n"""\nK個の1を用意して，これらを適当な個数に分割していくことを考える．\n合計でN個に分けたいので,これは玉を区別しない,箱を区別しない,各箱に最低1つ,\nの条件ぽいので分割数ぽい.\n\nただし,1を4個に分ける時に2パターンあったりする.\nn=?,K=1の時にいい感じの値を返すようにしたい.\n\n他にも，使う1の個数で場合分けというのもある．\n1をa個使うと，1以外のN-a個の数でK-aを作るということになり，これは値を2倍すればN-a個で2(K-a)を作るのと同値\n\n本番中は頭がバグっていたが，5個で1を作る時も3パターンあるね\nj=1の時だけ別個で求めようか．\n\n\nあー,dpテーブルを埋める時，i=jの時にdp[i][j]が1なんだから，jの降順に埋めてけばいいんじゃんか...\n斜め(桂馬っぽい動き)の累積和を持てば良さげ，だけど，この累積和の途中部分をどこかで利用しているのでそのマスをうまく使えば良い.\n\ndp[i][j]=dp[i][2j]+dp[i-1][j-1]\n"""\n\n\nN,K=MI()\nif N==K:\n    print((1))\n    return\n\nif K>N:\n    print((0))\n    return\n    \nM=max(N,K)\ndp=[[0]*(M+1) for _ in range(N+1)]\n# dp[i][j]はi個でjを作る\n\nfor i in range(1,M+1):\n    dp[i][i]=1\n\nfor i in range(1,N+1):\n    for j in range(i-1,0,-1):\n        dp[i][j]=dp[i-1][j-1]\n        if 2*j<=M:\n            dp[i][j]+=dp[i][2*j]\n        dp[i][j]%=mod\n\n\nprint((dp[N][K]))\n        \n# for i in range(N+1):\n#     print(dp[i])\n    \n'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: