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;

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	
	string str;
	int k;
	cin >> str >> k;
	int n=(int)str.size();
	long long mod=998244353;
	
	k=min(k,303);
	
	vector<int> a;
	int cccc=0;
	for(int i=0;i<n;i++){
		if(str[i]=='0'){
			a.push_back(cccc);
			cccc=0;
		}else{
			cccc++;
		}
	}
	a.push_back(cccc);
	n=(int)a.size();
	
	vector<vector<vector<long long>>> dp(n+1,vector<vector<long long>>(k+1,vector<long long>(k+1)));
	
	dp[0][0][0]=1;
	for(int i=0;i<n;i++){
		//subtract or nothing
		for(int j=0;j<=a[i];j++){
			for(int h=j;h<=k;h++){
				for(int s=0;s<=k;s++){
					if((h-j)>=0) dp[i+1][h-j][s]=(dp[i+1][h-j][s]+dp[i][h][s]);
					if(dp[i+1][h-j][s]>=mod) dp[i+1][h-j][s]-=mod;
				}
			}
		}
		//add
		for(int j=1;j<=k;j++){
			for(int h=0;h<=k;h++){
				if((h+j)>k) break;
				for(int s=0;s<=k;s++){
					if((s+j)>k) break;
					dp[i+1][h+j][s+j]+=dp[i][h][s];
					if(dp[i+1][h+j][s+j]>=mod) dp[i+1][h+j][s+j]-=mod;
				}
			}
		}
	}
	
	long long res=0;
	for(int i=0;i<=k;i++){
		res=(res+dp[n][0][i])%mod;
	}
	cout << res << "\n";
	return 0;
}