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<vector>
using namespace std;
const long mod=998244353;
string s;
int K;
long dp[303][303][303];
int main()
{
	cin>>s>>K;
	s+="0";
	if(K>s.size())K=s.size();
	vector<int>A;
	int cnt=0,sum=0;
	for(int i=0;i<s.size();i++)
	{
		if(s[i]=='1')cnt++;
		else
		{
			A.push_back(cnt);
			sum+=cnt;
			cnt=0;
		}
	}
	dp[0][sum][0]=1;
	for(int i=0;i<A.size();i++)
	{
		for(int j=0;j<=sum;j++)
		{
			int nAm=A[i]-(sum-j);
			if(nAm<0)nAm=0;
			for(int k=0;k<=K;k++)if(dp[i][j][k]>0)
			{
				dp[i][j][k]%=mod;
				int nAM=j;
				if(nAM>K-k+A[i])nAM=K-k+A[i];
				for(int l=nAm;l<A[i];l++)
				{
					dp[i+1][j-l][k]+=dp[i][j][k];
				}
				for(int l=nAm>A[i]?nAm:A[i],x=0;l<=nAM;l++)
				{
					dp[i+1][j-l][k+x]+=dp[i][j][k];
					x++;
				}
			}
		}
		sum-=A[i];
	}
	long ans=0;
	for(int k=0;k<=K;k++)(ans+=dp[A.size()][0][k])%=mod;
	cout<<ans<<endl;
}