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 <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <utility>
#include <tuple>
#include <cstdint>
#include <cstdio>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <deque>
#include <unordered_map>
#include <unordered_set>
#include <bitset>
#include <cctype>
#include <functional>
#include <ctime>
#include <cmath>
#include <limits>
#include <numeric>
#include <type_traits>
#include <iomanip>
#include <float.h>
#include <math.h>
using namespace std;
using ll = long long;


unsigned euclidean_gcd(unsigned a, unsigned b) {
    if (a < b) return euclidean_gcd(b, a);
    unsigned r;
    while ((r = a % b)) {
        a = b;
        b = r;
    }
    return b;
}

class UnionFind {
public:
    vector <ll> par;
    vector <ll> siz;
    UnionFind(ll sz_) : par(sz_), siz(sz_, 1LL) {
        for (ll i = 0; i < sz_; ++i) par[i] = i;
    }
    void init(ll sz_) {
        par.resize(sz_);
        siz.assign(sz_, 1LL);
        for (ll i = 0; i < sz_; ++i) par[i] = i;
    }
    ll root(ll x) {
        while (par[x] != x) {
            x = par[x] = par[par[x]];
        }
        return x;
    }
    bool merge(ll x, ll y) {
        x = root(x);
        y = root(y);
        if (x == y) return false;
        if (siz[x] < siz[y]) swap(x, y);
        siz[x] += siz[y];
        par[y] = x;
        return true;
    }

    bool issame(ll x, ll y) {
        return root(x) == root(y);
    }

    ll size(ll x) {
        return siz[root(x)];
    }
};

long long modpow(long long a, long long n, long long mod) {
    long long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}

long long modinv(long long a, long long mod) {
    return modpow(a, mod - 2, mod);
}

int main() {
    string s;
    ll k;
    cin >> s >> k;
    ll gd = 300;
    k = min(k, gd);
    vector<ll> z;
    ll now = 0;
    for (int i = 0; i < s.size(); i++) {
        if (s[i] == '1')now++;
        else {
            z.push_back(now);
            now = 0;
        }
    }
    z.push_back(now);
    reverse(z.begin(), z.end());
    vector<vector<ll>> dp(305,vector<ll>(k+1));
    dp[0][0] = 1;
    for (int i = 0; i < z.size(); i++) {
        vector<vector<ll>> zdp(305, vector<ll>(k+1));
        for (int j = 0; j < k+1; j++) {
            vector<ll> rui(305);
            for (int l = 0; l < 305; l++) {
                if (j + l > 300)break;
                if (j + l > k)break;
                if (l != 0)rui[l] += rui[l - 1];
                rui[l] += dp[l][j + l];
                if(l>=z[i]+1)rui[l] += 998244353-dp[l-z[i]-1][j + l-z[i]-1];
                rui[l]%= 998244353;
                zdp[l][j + l] += rui[l];
                zdp[l][j + l] += 998244353 - dp[l][j + l];
                zdp[l][j + l] %= 998244353;
            }
            vector<ll> r(305);
            for (int l = 302; l > -1; l--) {
                r[l] += r[l + 1];
                r[l] += dp[l][j];
                r[l] %= 998244353;
                zdp[l][j] += r[l];
                zdp[l][j] %= 998244353;
            }
        }
        for (int j = 0; j < 305; j++) {
            for (int n = 0; n < k+1; n++) {
                dp[j][n] = zdp[j][n];
            }
        }
    }
    ll ans = 0;
    for (int i = 0; i < k + 1; i++) {
        ans += dp[0][i];
    }
    cout << ans%998244353 << endl;
}