You are given a binary string (i. e. a string consisting of characters 0 and/or 1) s of length n. You can perform the following operation with the string s at most once: choose a substring (a contiguous subsequence) of s having exactly k characters 1 in it, and shuffle it (reorder the characters in the substring as you wish).

Calculate the number of different strings which can be obtained from s by performing this operation at most once.

Input

The first line contains two integers n and k (2 ≤ n ≤ 5000; 0 ≤ k ≤ n).

The second line contains the string s of length n, consisting of characters 0 and/or 1.

Output

Print one integer — the number of different strings which can be obtained from s by performing the described operation at most once. Since the answer can be large, output it modulo 998244353.

Examples

Input


7 2
1100110


Output


16


Input


5 0
10010


Output


1


Input


8 1
10001000


Output


10


Input


10 8
0010011000


Output


1

Note

Some strings you can obtain in the first example:

  * to obtain 0110110, you can take the substring from the 1-st character to the 4-th character, which is 1100, and reorder its characters to get 0110; 
  * to obtain 1111000, you can take the substring from the 3-rd character to the 7-th character, which is 00110, and reorder its characters to get 11000; 
  * to obtain 1100101, you can take the substring from the 5-th character to the 7-th character, which is 110, and reorder its characters to get 101. 



In the second example, k = 0 so you can only choose the substrings consisting only of 0 characters. Reordering them doesn't change the string at all, so the only string you can obtain is 10010.
5 2
00011
 5 4
11011
 81 5
100000101000000011001101010100001010111000000110000000001000101101111000001000110
 4 3
1011
 47 25
11110011111001110100101010011000110111011011101
 291 159
011001110111110110001111000110001001111100010001011000110100110010001101100110111010101111100111110000110111001100111010001000011001010011011100101000101011010110010011010100100101111010111000001011110101001011101110100100111110011001101010001110110101010101100011101110111000011111011001110
 459 15
101010101010101010101010101010101010101010101111111111111111111111010101010101010101010101011111111111111111111101010101010101010101111111111111111111111101010101010101010111111111111111111111111101010101010101111111111111111111111111110110101010101011111111111111111111111111111001000111111111111111111111111111110000011111111111111111111111111111000001111111111111111111111111111100001111111111111111111111111111100111111111111111111111111111110111111111111
 105 58
110101000111100000011011011101110100001000110011010101100010111001011111010010111101011011010110011011110
 52 11
0110011111001000010000100010010001010100000100110100

10
 5
 187538
 4
 314445058
 1
 0
 132732930
 478323746

import os
import sys
from io import BytesIO, IOBase
from types import GeneratorType
from collections import defaultdict
BUFSIZE = 8192
sys.setrecursionlimit(10**5)

def bootstrap(f,stack=[]):
    def fun(*args,**kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to=next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to=stack[-1].send(to)
            return to
    return fun


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")


# Python3 function to
# calculate nCr % p

# Python3 program to answer queries
# of nCr in O(1) time.
N = 100001

# array to store inverse of 1 to N
factorialNumInverse = [None] * (N + 1)

# array to precompute inverse of 1! to N!
naturalNumInverse = [None] * (N + 1)

# array to store factorial of
# first N numbers
fact = [None] * (N + 1)


# Function to precompute inverse of numbers
def InverseofNumber(p):
    naturalNumInverse[0] = naturalNumInverse[1] = 1
    for i in range(2, N + 1, 1):
        naturalNumInverse[i] = (naturalNumInverse[p % i] *
                                (p - int(p / i)) % p)


# Function to precompute inverse
# of factorials
def InverseofFactorial(p):
    factorialNumInverse[0] = factorialNumInverse[1] = 1

    # precompute inverse of natural numbers
    for i in range(2, N + 1, 1):
        factorialNumInverse[i] = (naturalNumInverse[i] *
                                  factorialNumInverse[i - 1]) % p


# Function to calculate factorial of 1 to N
def factorial(p):
    fact[0] = 1

    # precompute factorials
    for i in range(1, N + 1):
        fact[i] = (fact[i - 1] * i) % p


# Function to return nCr % p in O(1) time
def Binomial(N, R, p):
    # n C r = n!*inverse(r!)*inverse((n-r)!)
    ans = ((fact[N] * factorialNumInverse[R]) % p *
           factorialNumInverse[N - R]) % p
    return ans%p








p=998244353
InverseofNumber(p)
InverseofFactorial(p)
factorial(p)
t=1
for _ in range(t):
    #n=int(input())
    n,k=map(int,input().split())
    s=input()
    b=[]
    pre = [0]
    for j in s:
        b.append(int(j))
        pre.append(pre[-1]+b[-1])

    ans=1
    if(k>0):
        for i in range(n):
            j=i-1

            o=0
            z=0
            if(b[i]):
                o=1
            else:
                z=1

            while(j>=0):

                if((pre[i+1]-pre[j])>k):
                    j+=1
                    break
                else:
                    if(b[j]):
                        o+=1
                    else:
                        z+=1
                    j+=-1
            if(z>=1 and o>=1):
                if(b[i]):
                    z-=1
                else:
                    o-=1
                ans=(ans+(Binomial(z+o,z,p)))%p

    print(ans)











































































































































































