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 pb push_back
#define int long long
using namespace std;

const int N = 3e2 + 5;

int n, a[N], dp[N][N][N], pref[N];
const int M = 998244353;

signed main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    string S;
    int k;
    cin >> S >> k;
    int sz = S.size();
    k = min(k, sz);
    int cur = 0;
    for (int i = 0; i < S.size(); i++)
        if (S[i] == '1')
            cur++;
        else
            n++, a[n] = cur, cur = 0;
    n++;
    a[n] = cur;
    for (int i = 1; i <= n; i++)
        pref[i] = pref[i - 1] + a[i];
    dp[0][0][0] = 1;
    for (int i = 1; i <= n; i++)
        for (int s = pref[i]; s <= pref[n]; s++)
            for (int j = 0; j <= k; j++) {
                for (int q = max(1ll, a[i] - s); q <= a[i]; q++)
                    //if (s >= a[i] - q)
                        dp[i][s][j] += dp[i - 1][s - (a[i] - q)][j], dp[i][s][j] %= M;
                for (int q = 0; q <= min(j, s - a[i]); q++)
                   // if (s >= a[i] + q)
                        dp[i][s][j] += dp[i - 1][s - (a[i] + q)][j - q], dp[i][s][j] %= M;
                //cout << i << ' ' << s << ' ' << j << ' ' << dp[i][s][j] << endl;
            }
    int ans = 0;
    for (int j = 0; j <= k; j++)
        ans += dp[n][pref[n]][j], ans %= M;
    cout << ans;
    return 0;
}