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;
typedef long long ll;
const int N=305,mod=998244353;
int n,k,m,a[N],sum[N];
int dp[N][N][N];
//dp[i][j][h] 第i个坑，用了j个1，移动了k次的方案数
char s[N];
int main()
{
    scanf("%s%d",s+1,&k);
    for(int i=1;s[i];i++) m+=s[i]=='1';
    int t=0;
    for(int i=1;s[i];i++)
        if(s[i]=='0') a[++n]=t,t=0;
    else t++;
    a[++n]=t;
    for(int i=1;i<=n;i++) sum[i]=sum[i-1]+a[i];
    k=min(k,m);
    dp[0][0][0]=1;
    for(int i=0;i<n;i++)
        for(int j=sum[i];j<=sum[n];j++)
            for(int h=0;h<=k;h++)
        {
            for(int f=j;f<=sum[n];f++)
                if(f>=sum[i+1]&&h+f-j-a[i+1]<=k)
            {
                if(i+1==n&&f-j>a[i+1]) break;
                int cost=f-j<=a[i+1]?0:f-j-a[i+1];
                dp[i+1][f][h+cost]=(dp[i+1][f][h+cost]+dp[i][j][h])%mod;
            }
        }
    ll ans=0;
    for(int i=0;i<=k;i++) ans=(ans+dp[n][m][i])%mod;
    printf("%lld\n",ans);
}