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>
using namespace std;
typedef long long ll;
string s;
ll k,N,a[301],b[301],cnt,ans,dp[301][301][301],MOD=998244353;
int main(void){
    cin>>s>>k;
    for(int i=0;i<s.size();i++){
        if(s[i]-'0')a[N++]=cnt;
        else cnt++;
    }
    for(int i=0;i<N;i++){
        b[a[i]]++;
    }
    k=min(k,N);
    dp[cnt+1][0][0]=1;
    for(int i=cnt+1;i>=1;i--){
        for(int j=0;j<=k;j++){
            for(int l=0;l<=N;l++){
                if(dp[i][j][l]==0)continue;
                for(int x=0;x<=b[i-1]+l;x++){
                    dp[i-1][j+max(0LL,b[i-1]-x)][b[i-1]+l-x]+=dp[i][j][l];
                    dp[i-1][j+max(0LL,b[i-1]-x)][b[i-1]+l-x]%=MOD;
                }
            }
        }
    }
    for(int i=0;i<=k;i++){
        ans+=dp[0][i][0];
        ans%=MOD;
    }
    cout<<ans<<endl;
}