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;

const long long mo = 998244353;
const int maxi = 310;
long long dp[maxi][maxi][610];
int n, k;
string s;
int a[maxi];
int suf[maxi];
int main()
{
    cin>>s>>k;

    n = s.size();
    s = '#' + s;
    int zero = 0;
    int ones = 0;

    for (int i = 1;i<=n;i++)
        if (s[i] == '1') ones++; else
    {
        a[++zero] = ones;
        ones = 0;
    }

    a[zero + 1] = ones;

    suf[zero+1] = a[zero + 1];

    for (int i = zero;i>0;i--)
        suf[i] = suf[i+1] + a[i];

    k = min(k, n - zero);

    dp[0][0][0] = 1;


    for (int i = 1;i<=zero + 1;i++) //current group
        for (int j = 0; j<=k;j++) //performed operations
          for (int l = 0; l<=min(j, suf[i+1]); l++) //operations from suffix
          for (int w = -min(a[i], j); w<=l;w++){
                int o = max(0, w);
                dp[i][j][l]+=dp[i-1][j - o][l-w];
                if (dp[i][j][l]>=mo) dp[i][j][l]-=mo;
          }

   long long ans = 0;

   for (int i = 0;i <=k;i++)
   {
       ans+=dp[zero+1][i][0];
       if (ans>=mo) ans-=mo;
   }

   cout<<ans<<endl;
    return 0;
}