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;

const int maxn = 310, P = 998244353;
char s[maxn];
int K, f[maxn][maxn][maxn][2], sum[maxn][maxn];

int main() {
    scanf("%s %d", s + 1, &K);
    int n = strlen(s + 1), ans = 0;
    K = min(K, n);
    f[0][0][0][0] = 1;
    for (int i = 1; i <= n; i++) {
        memset(sum, 0, sizeof(s));
        for (int j = 0; j <= n; j++) {
            for (int k = 0; k <= K; k++) {
                sum[j][k] = f[i - 1][j][k][0];
                if (j) (sum[j][k] += sum[j - 1][k]) %= P;
            }
        }
        for (int j = 0; j <= n; j++) {
            for (int k = 0; k <= K; k++) {
                if (s[i] == '0') {
                    f[i][j][k][0] = (f[i - 1][j][k][0] + f[i - 1][j][k][1]) % P;
                    if (j) (f[i][j][k][0] += sum[j - 1][k]) %= P;
                } else {
                    f[i][j][k][0] = f[i - 1][j][k][0];
                    if (k) f[i][j][k][1] = (f[i - 1][j + 1][k - 1][0] + f[i - 1][j + 1][k - 1][1]) % P;
                }
            }
        }
    }
    for (int i = 0; i <= K; i++) {
        (ans += f[n][0][i][0]) %= P;
        (ans += f[n][0][i][1]) %= P;
    }
    printf("%d\n", ans);
    return 0;
}