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>

using namespace std;
#define int long long
const int MOD = 998244353;
int dp[301][301][301][2];
void add(int &a, const int &b) {
	a += b;
	if (a >= MOD) a -= MOD;
}
signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	string S;
	int K;
	cin >> S >> K;
	K = min(K, (int)300);
	int N = S.size();
	int res = 0;
	dp[0][0][0][0] = 1;
	for (int i = 1; i <= N; i++) {
		if (S[i - 1] == '0') {
			for (int j = 0; j <= K; j++) {
				for (int k = 0; k <= j; k++) {
					if (j > k) add(dp[i][j][k][0], dp[i][j - 1][k][0]);
					add(dp[i][j][k][0], dp[i - 1][j][k][0]);
				}
			}
			for (int j = 0; j <= K; j++) {
				for (int k = 0; k <= j; k++) {
					add(dp[i][j][k][0], dp[i - 1][j][k][1]);
				}
			}
		} else {
			for (int j = 0; j <= K; j++) {
				for (int k = 0; k <= j; k++) {
					if (k > 0) add(dp[i][j][k][1], dp[i - 1][j][k - 1][0]);
					if (k > 0) add(dp[i][j][k][1], dp[i - 1][j][k - 1][1]);

					//add(dp[i][j][k][1], dp[i - 1][j][k][1]);
					add(dp[i][j][k][0], dp[i - 1][j][k][0]);
				}
			}
		}
	}
	
	for (int i = 0; i <= K; i++) {
		//cerr << i << " " << dp[N][i][i][0] << " " << dp[N][i][i][1] << endl;
		add(res, dp[N][i][i][0]);
		add(res, dp[N][i][i][1]);
	}
	cout << res << endl;
}