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)

#include <iostream>
#include <vector>
#include <array>
#include <list>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <string>
#include <sstream>
#include <algorithm>
#include <random>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <cmath>
#include <cassert>
#include <climits>
#include <bitset>
#include <functional>
#include <iomanip>
#include <random>

#define FOR_LT(i, beg, end) for (int i = (int)(beg); i < (int)(end); i++)
#define FOR_LE(i, beg, end) for (int i = (int)(beg); i <= (int)(end); i++)
#define FOR_DW(i, beg, end) for (int i = (int)(beg); (int)(end) <= i; i--)
#define REP(n)              for (int repeat_index = 0; repeat_index < (int)n; repeat_index++)


using namespace std;

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout << fixed << setprecision(20);

	int64_t v; cin >> v; v--;

	string s;
	while (v >= 0) {
		s.push_back('a' + (v % 26));
		v /= 26;
		v--;
	}

	string t;
	FOR_LT(i, 0, s.size()) {
		t.push_back(s[s.size() - 1 - i]);
	}
	cout << t << endl;

	return 0;
}