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;
#define oo 1000000010
#define mod 998244353
const int N = 310;

char s[N];

int n , k;

vector< int > v;

int dp[N][N][N];

int solve(int i,int give,int used){
	if(used > k)
		return 0;
	if(i == (int)v.size()){
		if(give > 0)
			return 0;
		return 1;
	}
	if(dp[i][give][used] != -1)
		return dp[i][give][used];
	dp[i][give][used] = solve(i + 1 , give, used);
	for(int j = 1;j <= give;j++){
		dp[i][give][used] += solve(i + 1 , give - j , used);
		if(dp[i][give][used] >= mod) dp[i][give][used] -= mod;
	}
	for(int j = 1;j <= v[i];j++){
		dp[i][give][used] += solve(i + 1 , give + j , used + j);
		if(dp[i][give][used] >= mod) dp[i][give][used] -= mod;
	}
	return dp[i][give][used];
}

int main(){
	scanf("%s",s);
	n = strlen(s);
	scanf("%d",&k);
	int cnt = 0;
	for(int i = 0 ;i < n;i++){
		if(s[i] == '0'){
			v.push_back(cnt);
			cnt = 0;
		}
		else
			cnt++;
	}
	v.push_back(cnt);
	reverse(v.begin(),v.end());
	memset(dp,-1,sizeof(dp));
	cout << solve(0 , 0 , 0) << endl;
	return 0;
}