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++ (Clang 10.0.0) (judge verdict: Accepted)

#pragma GCC optimize ("O2")
#pragma GCC target ("avx2")
//#include<bits/stdc++.h>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i = 0; i < (n); i++)
#define rep1(i, n) for(int i = 1; i <= (n); i++)
#define co(x) cout << (x) << "\n"
#define cosp(x) cout << (x) << " "
#define ce(x) cerr << (x) << "\n"
#define cesp(x) cerr << (x) << " "
#define pb push_back
#define mp make_pair
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
#define Would
#define you
#define please

const ll mod = 998244353;

ll dp[310][310];

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);


	string S;
	int K;
	int k = 0;
	int A[301] = {};

	cin >> S >> K;
	rep(i, S.size()) {
		if (S[i] == '0') k++;
		else A[k]++;
	}

	int kosuu = 0;
	dp[0][0] = 1;
	for (int i = k; i >= 0; i--) {
		if ((i & 3) == 0) {
			for (int j = kosuu; j >= 0; j--) {
				ll tmp = 0;
				for (int k = kosuu; k >= 0; k--) {
					dp[j][k] %= mod;
					tmp += dp[j][k];
					rep1(p, A[i]) dp[j + p][k + p] += dp[j][k];
					dp[j][k] = tmp;
				}
			}
		}
		else {
			for (int j = kosuu; j >= 0; j--) {
				ll tmp = 0;
				for (int k = kosuu; k >= 0; k--) {
					tmp += dp[j][k];
					rep1(p, A[i]) dp[j + p][k + p] += dp[j][k];
					dp[j][k] = tmp;
				}
			}
		}
		kosuu = min(kosuu + A[i], K);		
	}

	ll kotae = 0;
	rep(i, min(K, kosuu) + 1) {
		kotae += dp[i][0] % mod;
	}
	co(kotae % mod);



	Would you please return 0;
}