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 <bits/stdc++.h>
#define all(x) x.begin(), x.end()
#define sz(x) (int)x.size()
#define x first
#define y second
using namespace std;

const int N = 302;
const int M = 998244353;

string s;
int n = 0;
int counter = 0;
int ones[N];
int dp[N][N][N]; // prefix, debt, sum

signed main() {
#ifdef LC
    assert(freopen("input.txt", "r", stdin));
#endif
    ios::sync_with_stdio(0), cin.tie(0);

	cin >> s;
	int debt;
	cin >> debt;
	s += "0";
	for (char c : s) {
		if (c == '1') {
			++counter;
		} else {
			ones[++n] = counter;
			counter = 0;
		}
	}
	//cerr << "n = " << n << endl;
	//for (int i = 1; i <= n; ++i) {
	//	cerr << ones[i] << " ";
	//}
	//cerr << endl;
	dp[0][0][0] = 1;
	int total = count(all(s), '1');
	debt = min(debt, total);
	//cerr << "d t " << debt << " " << total << endl;
	for (int i = 1; i <= n; ++i) {
		for (int j = 0; j <= total; ++j) {
			for (int k = j; k <= debt; ++k) {
				for (int d = 0; d <= j; ++d) {
					(dp[i][j][k] += dp[i - 1][j - d][k - d]) %= M;
				}
				for (int d = 1; d <= ones[i]; ++d) {
					(dp[i][j][k] += dp[i - 1][j + d][k]) %= M;
				}
			}
		}
	}
	int ans = 0;
	for (int i = 0; i <= debt; ++i) {
		(ans += dp[n][0][i]) %= M;
	}
	cout << ans << "\n";
    return 0;
}