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;
typedef unsigned long long ull;
typedef pair<int,int> pii;
#define x first
#define y second
#define mp make_pair
#define pb push_back
template <typename TYPE> inline void chkmax(TYPE &x,TYPE y){x<y?x=y:0;}
template <typename TYPE> inline void chkmin(TYPE &x,TYPE y){y<x?x=y:0;}
template <typename TYPE> void readint(TYPE &x)
{
    x=0;int f=1;char c;
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-1;
    for(;isdigit(c);c=getchar())x=x*10+c-'0';
    x*=f;
}
const int MAXN=305,MOD=998244353;
inline int dmy(int x){return x>=MOD?x-MOD:x;}

int n,m,P;
char s[MAXN];
int a[MAXN],f[MAXN][MAXN][MAXN],p[MAXN][MAXN][MAXN],q[MAXN][MAXN][MAXN];

int main()
{
	#ifdef LOCAL
	freopen("code.in","r",stdin);
//	freopen("code.out","w",stdout);
	#endif
	scanf("%s",s+1);readint(P);
	n=strlen(s+1);
	m=1;
	for(int i=1;i<=n;++i)
		if(s[i]=='0')++m;
		else ++a[m];
	f[0][0][0]=1;
	for(int j=0;j<=n;++j)p[0][j][0]=q[0][j][j]=1;
	for(int i=1;i<=m;++i)
		for(int j=0;j<=n;++j)
			for(int k=0;k<=n;++k)
			{
				f[i][j][k]=dmy(dmy(p[i-1][min(n,j+a[i])][k]-p[i-1][j][k]+MOD)+q[i-1][j][k]);
				if(j)p[i][j][k]=dmy(p[i][j-1][k]+f[i][j][k]);
				else p[i][j][k]=f[i][j][k];
				if(j && k)q[i][j][k]=dmy(q[i][j-1][k-1]+f[i][j][k]);
				else q[i][j][k]=f[i][j][k];
			}
	int ans=0;
	for(int k=0;k<=n && k<=P;++k)
		ans=dmy(ans+f[m][0][k]);
	printf("%d\n",ans);
	return 0;
}