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;

int N, M, K;
char s[310];
int a[310];
int f[310][310][310];
const int MOD = 998244353;
void upd(int &x, int y) {
    x += y;
    if (x >= MOD) x -= MOD;
}

int main() {
    scanf("%s", s);
    scanf("%d", &K);
    N = strlen(s);
    int tmp = 0;
    M = 0;
    for (int i=0;i<N;i++) {
        if (s[i] == '1') {
            tmp++;
        } else {
            a[++M] = tmp;
            tmp = 0;
        }
    }
    a[++M] = tmp;
    //for (int i=1;i<=M;i++) printf("%d %d\n",i,a[i]);
    K = min(K, N);
    f[0][0][0] = 1;
    for (int i=0;i<=M;i++) {
        for (int j=0;j<=K;j++) {
            for (int k=0;k<=K;k++) {
                int v = f[i][j][k];
                if (!v) continue;
                assert(j<=k);
                // return
                for (int l=0;l<=j && l<=a[i+1];l++) {
                    upd(f[i+1][j-l][k], v);
                }
                // borrow
                for (int l=1;k+l<=K;l++) {
                    upd(f[i+1][j+l][k+l], v);
                }
            }
        }
    }
    int ans = 0;
    for (int i=0;i<=K;i++) {
        upd(ans, f[M][0][i]);
    }
    printf("%d\n",ans);
}