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<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int N=305;
const int MOD=998244353;
int n,m;
char s[N];
int a[N];
int sum[N];
long long dp[N][N][N];
int main()
{
    scanf("%s%d",s+1,&m);
    n=strlen(s+1);
    s[++n]='0';
    int cnt=0;
    for(int i=1;i<=n;i++) cnt+=s[i]-'0';
    m=min(m,cnt);
    cnt=0;
    for(int i=1;i<=n;i++)
        if(s[i]=='0') cnt++;
        else a[cnt+1]++;
    for(int i=1;i<=cnt;i++)
        sum[i]=sum[i-1]+a[i];
    dp[0][0][0]=1;
    for(int i=1;i<=cnt;i++)
        for(int j=sum[i];j<=n-cnt;j++)
            for(int k=0;k<=m;k++)
                for(int p=0;p<=min(j,n-sum[i-1]-cnt);p++)
                    if(k>=max(p-a[i],0)) dp[i][j][k]=(dp[i][j][k]+dp[i-1][j-p][k-max(p-a[i],0)])%MOD;
    long long ans=0;
    for(int k=0;k<=m;k++)
        ans=(ans+dp[cnt][n-cnt][k])%MOD;
    printf("%lld",ans);
    return 0;
}