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;
char s[500];
int K,N,cnt,a[500],f[2][610][610];
const int P = 998244353;
int main(){
	scanf("%s%d",s+1,&K);
	N = strlen(s+1);
	cnt = 1;
	int sum = 0;
	for (int i=1; i<=N; i++){
		if (s[i] == '1') a[cnt]++;
		else cnt++,sum += a[cnt-1];
	}
	if (!a[cnt]) cnt--;
//	printf("%d\n",cnt);
//	for (int i=1; i<=cnt; i++) printf("%d ",a[i]); puts("");
	K = min(K,N)<<1;
	f[0][0][0] = 1;
	for (int i=0,op=0; i<cnt; i++,op^=1){
		memset(f[op^1],0,sizeof(f[op^1]));
		for (int j=0; j<=K; j++){
			for (int k=0; k<=N; k++){
				if (!f[op][j][k]) continue;
//				printf("debug %d %d %d %d\n",i,j,k,f[i][j][k]);
				f[op^1][j][k] += f[op][j][k];
				f[op^1][j][k] %= P;
				for (int l=1; j+l<=K; l++)
					f[op^1][j+l][k+l] += f[op][j][k], f[op^1][j+l][k+l] %= P;
				for (int l=1; l<=min(a[i+1],k) && j+l<=K; l++)
					f[op^1][j+l][k-l] += f[op][j][k], f[op^1][j+l][k-l] %= P; 
			}
		}	
	}
	int ans = 0;
	for (int i=0; i<=K; i++) ans += f[cnt%2][i][0], ans %= P;
	printf("%d\n",ans);
	return 0;
}