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>
#include<queue>
#include<cmath>
#include<bitset>
#include<map>
#include<vector>
using namespace std;
typedef long long ll;
const int N=310,mod=998244353;
char s[N];
int K,ct,n,m;
int w[N];
ll sum,f[N][N][N];
void pls(ll &x,ll y){x+=y;x>=mod?x-=mod:0;}
int main(){
//	freopen("a.in","r",stdin);
	scanf("%s%d",s+1,&K);n=strlen(s+1);
	for(int i=1;i<=n;i++)if(s[i]=='0')m++,w[m]=ct,ct=0;else ct++;
	m++;w[m]=ct;
	f[m+1][0][0]=1;
	for(int i=m;i;i--){sum=min(sum,(ll)K);
		for(int j=0;j<=sum;j++)for(int k=j;k<=sum;k++)for(int c=0;c<=j+w[i];c++){
			if(c<w[i])pls(f[i][j+w[i]-c][k+w[i]-c],f[i+1][j][k]);
			else pls(f[i][j+w[i]-c][k],f[i+1][j][k]);
		}
		sum+=w[i];
	}
	K=min(K,n);
	sum=0;for(int i=0;i<=K;i++)pls(sum,f[1][0][i]);
	cout<<sum;
}