Problem:
Score : 800 points

Problem Statement

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).

Sample Input 1

0101 1

Sample Output 1

4

Four strings, 0101, 0110, 1001, and 1010, can result.

Sample Input 2

01100110 2

Sample Output 2

14

Sample Input 3

1101010010101101110111100011011111011000111101110101010010101010101 20

Sample Output 3

113434815

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#define _USE_MATH_DEFINES
#define _SILENCE_CXX17_C_HEADER_DEPRECATION_WARNING
#define _SILENCE_ALL_CXX17_DEPRECATION_WARNINGS
#define _CRT_SECURE_NO_WARNINGS
//#include <bits/stdc++.h>

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
#include <cassert>
#include <fstream>
#include <array>
#include <string>
#include <cstring>
#include <functional>
#include <ctime>
#include <iostream>
#include <unordered_set>
#include <complex>

#include <nmmintrin.h>
#include <mmintrin.h>
#include <emmintrin.h>


using namespace std;

#define dprint(Exp,...) if(Exp){fprintf(stderr, __VA_ARGS__);}
#define printe(...) fprintf(stderr, __VA_ARGS__);
//#define printe(...) void();
#define printE(...) fprintf(stderr, __VA_ARGS__);
//#define printE(...) void();


#define PrtExp(_Exp)  cerr<< #_Exp <<" = "<< (_Exp)
#define PrtExpN(_Exp)  cerr<< #_Exp <<" = "<< (_Exp) <<"\n"

#define SINT(n) scanf("%d",&n)
#define SINT2(n,m) scanf("%d %d",&n,&m)
#define SINT3(n,m,o) scanf("%d %d %d",&n,&m,&o)
#define SINT4(n,m,o,P) scanf("%d %d %d %d",&n,&m,&o,&P)
#define SINT5(n,m,o,P,q) scanf("%d %d %d %d %d",&n,&m,&o,&P,&q)
#define SLL(n) scanf("%lld",&n)
#define SLL2(n,m) scanf("%lld %lld",&n,&m)
#define SLL3(n,m,o) scanf("%lld %lld %lld",&n,&m,&o)
#define SST(s) scanf("%s",s)
#define SCH(c) scanf("%c",&c)

#define GC() getchar()

#define PINT(n) printf("%d",(int)(n))
#define PINT2(n,m) printf("%d %d",(int)(n),(int)(m))
#define PINT3(n,m,l) printf("%d %d %d",(int)(n),(int)(m),(int)(l))
#define PLL(n) printf("%lld",(long long)(n))
#define PST(s) printf("%s",(s))
#define PCH(s) printf("%c",(s))

#define PINTN(n) printf("%d\n",(int)(n))
#define PINT2N(n,m) printf("%d %d\n",(int)(n),(int)(m))
#define PINT3N(n,m,l) printf("%d %d %d\n",(int)(n),(int)(m),(int)(l))
#define PLLN(n) printf("%lld\n",(long long)(n))
#define PSTN(s) printf("%s\n",(s))
#define PCHN(s) printf("%c\n",(s))

#define PSP() printf(" ")
#define PN() printf("\n")

#define PC(c) putchar(c)
#define CSP (' ')
#define SN ("\n")

#define rep(i,a) for(int i=0;i<a;i++)
#define reP(i,a) for(int i=0;i<=a;i++)
#define Rep(i,a) for(int i=a-1;i>=0;i--)
#define ReP(i,a) for(int i=a;i>=0;i--)

#define rEp(i,a) for(i=0;i<a;i++)
#define rEP(i,a) for(i=0;i<=a;i++)
#define REp(i,a) for(i=a-1;i>=0;i--)
#define REP(i,a) for(i=a;i>=0;i--)

#define repft(i,a,b) for(int i=a;i<b;i++)
#define repfT(i,a,b) for(int i=a;i<=b;i++)
#define Repft(i,a,b) for(int i=a-1;i>=b;i--)
#define RepfT(i,a,b) for(int i=a;i>=b;i--)

#define foreach(a,it) for(auto it = a.begin(); it != a.end(); ++it)

#define FILL(a,v) fill(begin(a),end(a), v)
#define FILL0(a) memset(a,0,sizeof(a))
#define FILL1(a) memset(a,-1,sizeof(a))

typedef long long ll;

typedef unsigned long long ull;
typedef pair<int, int> Pi;
typedef pair<ll, ll>   Pll;


const int INF = 1'010'000'000; // 0x3C33'6080
const ll INFLL = 0x1f1f1f1f1f1f1f1fLL;//2,242,545,357,980,376,863


template <class A, class B> inline ostream& operator<<(ostream& st, const pair<A, B>& P) { return st << "(" << P.first << "," << P.second << ")"; };
template <class A, class B> inline pair<A, B> operator+(const pair<A, B>& P, const pair<A, B>& Q) { return pair<A, B>(P.first + Q.first, P.second + Q.second); };
template <class A, class B> inline pair<A, B> operator-(const pair<A, B>& P, const pair<A, B>& Q) { return pair<A, B>(P.first - Q.first, P.second - Q.second); };


inline ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }

int bitcount(long bits) {
    bits = (bits & 0x55555555) + (bits >> 1 & 0x55555555);
    bits = (bits & 0x33333333) + (bits >> 2 & 0x33333333);
    bits = (bits & 0x0f0f0f0f) + (bits >> 4 & 0x0f0f0f0f);
    bits = (bits & 0x00ff00ff) + (bits >> 8 & 0x00ff00ff);
    return (bits & 0x0000ffff) + (bits >> 16 & 0x0000ffff);
}

vector<int> conv(const string s) {
    vector<int> ret;
    ret.reserve(s.size());
    int a = 0;
    for (const char c : s) {
        if (c == '0') {
            ret.push_back(a);
            a = 0;
        } else {
            ++a;
        }
    }
    ret.push_back(a);
    return ret;
}

const int N = 500;

ll dpa[N+2][N+2][N+2];

const ll M = 998244353;

int main() {
    string s;
    int k;
    cin >> s >> k;


    auto v = conv(s);
    reverse(v.begin(), v.end());
    //for (auto c : v) cout << c << " "; cout << endl;

    FILL0(dpa);
    dpa[0][0][0] = 1;
    // dp[i][a][b]
    // a : sum stock
    // b : now stock

    const int n = v.size();
    k = min({ k, n * 3, N });
    rep(i, n) {
        int c = v[i];
        reP(a, k) {
            reP(b, k) {
                reP(d, c) {
                    if (a + d > k) break;
                    if (b + d > k) break;
                    //(dpa[i + 1][a + d][b + d] += dpa[i][a][b]) %= M;
                    dpa[i + 1][a + d][b + d] += dpa[i][a][b];
                }
                repfT(d, 1, b) {
                    //(dpa[i + 1][a][b - d] += dpa[i][a][b]) %= M;
                    dpa[i + 1][a][b - d] += dpa[i][a][b];
                }
            }
        }
        reP(a, k) {
            reP(b, k) {
                dpa[i + 1][a][b] %= M;
            }
        }
    }
    ll ret = 0;
    reP(i, k) {
        ret += dpa[n][i][0];
        ret %= M;
    }


    cout <<ret << endl;


    return 0;
}


/*
01100110 : 27
10110010
11011000 : 64


01100110

01101100

*/