Problem: Given is a string S consisting of `0` and `1`. Find the number of strings, modulo 998244353, that can result from applying the following operation on S between 0 and K times (inclusive):

* Choose a pair of integers i, j (1\leq i < j\leq |S|) such that the i-th and j-th characters of S are `0` and `1`, respectively. Remove the j-th character from S and insert it to the immediate left of the i-th character.

Constraints

* 1 \leq |S| \leq 300
* 0 \leq K \leq 10^9
* S consists of `0` and `1`.

Input

Input is given from Standard Input in the following format:


S K


Output

Find the number of strings, modulo 998244353, that can result from applying the operation on S between 0 and K times (inclusive).

Examples

Input

0101 1


Output

4


Input

01100110 2


Output

14


Input

1101010010101101110111100011011111011000111101110101010010101010101 20


Output

113434815

Solution (in PYTHON3):

Solution:

S,K = input().split()
S = list(S)

ori = [0]
for i in range(len(S)-1,-1,-1):
    if S[i] == "0":
        ori.append(0)
    else:
        ori[-1] += 1

#print (ori)
N = len(ori)

K = int(K)
K = min(310,K)
mod = 998244353

dp = [[[0] * (K+1) for i in range(K+1)] for j in range(N+10)]

dp[-1][0][0] = 1


for i in range(N):

    #なにもしない
    for k in range(K+1):
        for x in range(k+1):
            dp[i][k][x] += dp[i-1][k][x]

    #1を追加する
    for x in range(K+1):

        nsum = 0

        for k in range(K+1):

            if x <= k:
                dp[i][k][x] += nsum
                dp[i][k][x] %= mod

            nsum += dp[i-1][k][x]

            if k-ori[i] >= 0:
                nsum -= dp[i-1][k-ori[i]][x]
            nsum %= mod

    #xを使用する

    for k in range(K+1):

        nsum = 0

        for x in range(k+1):

            dp[i][k][x] += nsum
            dp[i][k][x] %= mod

            if x >= 0:
                nsum += dp[i-1][k][x]

        nsum %= mod

    #print (dp[i])

ans = 0
for i in range(K+1):
    ans += dp[N-1][i][i]
    ans %= mod
print (ans)