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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#define Maxn 307
#define modp 998244353
int len,n,K;
int a[Maxn];
int f[Maxn][Maxn][Maxn],s1[Maxn][Maxn][Maxn],s2[Maxn][Maxn][Maxn];
char s[Maxn];
int main()
{
    scanf("%s%d",s+1,&K);
    len=strlen(s+1);
    n=0;
    for (int i=1;i<=len;i++)
        if (s[i]=='1') ++a[n]; else ++n;
    if (K>len) K=len;
    f[n+1][0][0]=1;
    for (int i=0;i<=len;i++)
        s1[n+1][i][0]=1,s2[n+1][i][i]=1;
    int sum=0;
    for (int i=n;i>=0;i--)
    {
        sum+=a[i];
        for (int j=0;j<=sum;j++)
            for (int k=0;k<=K;k++)
            {
                f[i][j][k]=s1[i+1][j][k];
                if (j-a[i]>0) f[i][j][k]=(f[i][j][k]+modp-s1[i+1][j-a[i]-1][k])%modp;
                int lx=a[i]+1,rx=min(j,k+a[i])+1;
                if (lx<rx) f[i][j][k]=(f[i][j][k]+s2[i+1][j-lx][k+a[i]-lx])%modp;
                s1[i][j][k]=f[i][j][k];
                if (j>0) s1[i][j][k]=(s1[i][j][k]+s1[i][j-1][k])%modp;
                s2[i][j][k]=f[i][j][k];
                if (j>0&&k>0) s2[i][j][k]=(s2[i][j][k]+s2[i][j-1][k-1])%modp;
            }
        for (int j=sum+1;j<=len;j++)
            for (int k=0;k<=K;k++)
            {
                f[i][j][k]=0;
                if (j>0) s1[i][j][k]=s1[i][j-1][k];
                if (j>0&&k>0) s2[i][j][k]=s2[i][j-1][k-1];
            }
    }
    int ans=0;
    for (int i=0;i<=K;i++)
        ans=(ans+f[0][sum][i])%modp;
    printf("%d\n",ans);
    return 0;
}