Problem: 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).

Examples

Input

0101 1


Output

4


Input

01100110 2


Output

14


Input

1101010010101101110111100011011111011000111101110101010010101010101 20


Output

113434815

Solution (in CPP):

Solution:
#include <iostream>
#include <vector>
#include <cstring>

using namespace std;

#define forsn(i,s,n) for(int i=s;i<(int)(n);i++)
#define forn(i,n) forsn(i,0,n)

typedef long long int tint;

const tint MOD = 998244353;

int balanced;
vector<int> pos;

int memo[310][610][310];
tint f(int idx, int balance, int missingOps) {
	if (idx + 1 == (int) pos.size()) return balance == balanced;
	if (memo[idx][balance][missingOps] != -1) return memo[idx][balance][missingOps];
	
	tint res = f(idx + 1, balance, missingOps);
	if (balance > balanced) {
		forsn(j, 1, min(balance - balanced + 1, pos[idx + 1] - pos[idx])) {
			res += f(idx + 1, balance - j, missingOps);
			res %= MOD;
		}
	}
	forsn(i, 1, missingOps + 1) {
		res += f(idx + 1, balance + i, missingOps - i);
		res %= MOD;
	}
	
	memo[idx][balance][missingOps] = res;
	return res;
}

int main() {
	memset(memo, -1, sizeof(memo));
	
	string s; cin >> s;
	int k; cin >> k; 
	k = min(k, (int)s.size());
	
	balanced = s.size();
	
	pos.push_back(-1);
	int starting = 0; while (starting < (int)s.size() && s[starting] != '0') starting++;
	if (starting < (int)s.size()) pos.push_back(0);
	forsn(i, 1, s.size() - starting) if (s[starting + i] == '0') pos.push_back(i);
	if (starting < (int)s.size()) pos.push_back(s.size() - starting);
	
	cout << f(0, balanced, k) << endl;
}