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;
using ll = long long;

const int N = 303;
ll dp[N][N][N];
vector<int> v;
const ll M = 998244353LL;

int main() {
    string s; cin >> s;
    v.push_back(0);
    int c = 0; cin >> c;
    for (int i = 0; i < s.size(); i++) {
        if (s[i] == '1') {
            v.back() += 1;
        } else {
            v.push_back(0);
        }
    }
    c = min(c, accumulate(begin(v), end(v), 0));
    reverse(begin(v), end(v));
    dp[0][0][0] = 1;
    for (int i = 0; i < v.size()-1; i++) {
        for (int j = 0; j <= c; j++) {
            for (int k = 0; k <= c; k++) {
                for (int l = 0; l <= k; l++) {
                    dp[i+1][j][k-l] += dp[i][j][k];
                    dp[i+1][j][k-l] %= M;
                }
                for (int l = 1; l <= min(v[i], min(c-j, c-k)); l++) {
                    dp[i+1][j+l][k+l] += dp[i][j][k];
                    dp[i+1][j+l][k+l] %= M;
                }
            }
        }
    }
    int ans = 0;
    for (int i = 0; i <= c; i++) {
        for (int j = 0; j <= c; j++) {
            ans += dp[v.size()-1][i][j];
            ans %= M;
        }
    }
    cout << ans << endl;

    return 0;
}