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: Accepted)

#include <bits/stdc++.h>
using namespace std;

using ll = long long;
using Pi = pair<int, int>;
using Pl = pair<ll, ll>;
using vint = vector<int>;
using vvint = vector<vint>;
using vvvint = vector<vvint>;
using vdouble = vector<double>;
using vvdouble = vector<vdouble>;
using vvvdouble = vector<vvdouble>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vvvll = vector<vvll>;
using uint = unsigned int;
using ull = unsigned long long;

template<typename T> using uset = unordered_set<T>;
template<typename T1, typename T2> using umap = unordered_map<T1, T2>;

constexpr int INF = (1 << 30) - 1;
constexpr ll LLINF = 1LL << 60;
constexpr int dy[] = {1, 0, -1, 0, 1, -1, -1, 1};
constexpr int dx[] = {0, 1, 0, -1, 1, 1, -1, -1};
constexpr char el = '\n';
constexpr int mod = 1000000007;
constexpr int mod2 = 998244353;
template<typename T> T gcd(T a, T b) { return (b ? gcd(b, a % b) : a); }
template<typename T> T lcm(T a, T b) { return (a / gcd(a, b) * b); }
template<typename T1, typename T2>
inline bool chmin(T1 &a, T2 b) { return (a > b && (a = b, true)); }
template<typename T1, typename T2>
inline bool chmax(T1 &a, T2 b) { return (a < b && (a = b, true)); }

template<typename T>
vector<T> makeVector(size_t a, T b) { return (vector<T>(a, b)); }
template<typename... Ts>
auto makeVector(size_t a, Ts... ts) { 
	return (vector<decltype(makeVector(ts...))>(a, makeVector(ts...)));
}

template<typename T>
bool isin(T y, T x, T h, T w) { return (0 <= y && 0 <= x && y < h && x < w); }
template<typename T>
bool isin1(T y, T x, T h, T w) { return (0 < y && 0 < x && y <= h && x <= w); }

template<typename T>
ostream& operator <<(ostream &os, vector<T> &v) {
	os << v[0];
	for (int i = 1; i < v.size(); i++) os << " " << v[i];
	return (os);
}

template<typename T>
istream& operator >>(istream &is, vector<T> &v) {
	for (auto &u : v) is >> u;
	return (is);
}

template<typename T1, typename T2>
istream& operator >>(istream &is, pair<T1, T2> &p) {
	return (is >> p.first >> p.second);
}


void Main() {
	string S;
	int K;
	cin >> S >> K;
	K = min(K, 300);
	
	int cnt = 0;
	vint A;
	for (int i = 0; i < S.size(); i++) {
		if (S[i] == '1') {
			cnt++;
		} else {
			A.push_back(cnt);
			cnt = 0;
		}
	}
	A.push_back(cnt);
	
	reverse(begin(A), end(A));
	//cout << A << endl;

	int N = A.size();
	auto dp = makeVector(N+1, 305, K+2, 2, 0ll);
	
	dp[0][0][K][0] = 1;
	dp[0][1][K-1][0] = -1;
	for (int i = 0; i < N; i++) {
		for (int j = 1; j <= 300; j++) {
			for (int k = K; k >= 0; k--) {
				(dp[i][j][k][0] += dp[i][j-1][k+1][0]) %= mod2;
			}
		}

		for (int j = 300; j >= 0; j--) {
			for (int k = K; k >= 0; k--) {
				if (j > 0) {
					(dp[i][j-1][k][1] += dp[i][j][k][0] + dp[i][j][k][1]) %= mod2;
				}

				(dp[i+1][j][k][0] += dp[i][j][k][0]) %= mod2;
				if (j+A[i]+1 <= 300 && k-A[i]-1 >= 0) (dp[i+1][j+A[i]+1][k-A[i]-1][0] += mod2 - dp[i][j][k][0]) %= mod2;

				(dp[i+1][j][k][0] += dp[i][j][k][1]) %= mod2;
				if (j+1 <= 300 && k > 0) {
					(dp[i+1][j+1][k-1][0] += mod2 - dp[i][j][k][1]) %= mod2;
				}
			}
		}
	}

	ll sum = 0;
	for (int i = 0; i <= K; i++) {
		(sum += dp[N][0][i][0] + dp[N][0][i][1]) %= mod2;
	}
	cout << sum << endl;
}

int main() {
	cin.tie(0);
	ios::sync_with_stdio(0);
	cout << fixed << setprecision(20);
	Main();
	return (0);
}