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 <iostream>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <random>
#include <iomanip>
#include <functional>
#include <cassert>

using namespace std;

typedef long long ll;

const int md = 998244353;

void add(int &a, int b) {
  a += b;
  if (a >= md) a -= md;
}

const int N = 307;

int dp[N][N][N];

int main() {
  ios_base::sync_with_stdio(false); cin.tie(0);
#ifdef LOCAL
  freopen("input.txt", "r", stdin);
#endif
  string s; int k;
  cin >> s >> k;
  int n = s.size();
  vector <int> have;
  int cur = 0;
  for (int i = 0; i < n; ++i) {
    if (s[i] == '0') {
      have.push_back(cur);
      cur = 0;
    } else {
      ++cur;
    }
  }
  have.push_back(cur);
  int m = have.size();
  dp[m][0][0] = 1;
  for (int i = m - 1; i >= 0; --i) {
    for (int go = 0; go <= n; ++go) {
      for (int cost = 0; cost <= n; ++cost) {
        if (dp[i + 1][go][cost] == 0) continue;
        for (int d = 0; d <= go; ++d) {
          add(dp[i][go - d][cost], dp[i + 1][go][cost]);
        }
        for (int d = 1; d <= have[i]; ++d) {
          add(dp[i][go + d][cost + d], dp[i + 1][go][cost]);
        }
      }
    }
  }
  int ans = 0;
  for (int i = 0; i <= min(k, N - 1); ++i) {
    add(ans, dp[0][0][i]);
  }
  cout << ans << '\n';
}