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 MOD = 998244353;

int K;
vector<int> ones;
int dp[305][305][305];

int f(int pos, int ops, int cnt) {
  if (ops > K || cnt < 0) return 0;
  if (pos < 0) return cnt == 0;
  int &res = dp[pos][ops][cnt];
  if (res != -1) return res;
  res = 0;
  for (int i = 0; i <= ones[pos]; i++) {
    res += f(pos - 1, ops + i, cnt + i);
    if (res >= MOD) res -= MOD;
  }
  for (int i = 1; i <= cnt; i++) {
    res += f(pos - 1, ops, cnt - i);
    if (res >= MOD) res -= MOD;
  }
  return res;
}

int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  memset(dp, -1, sizeof(dp));

  string S;
  cin >> S >> K;
  ones.push_back(0);
  for (auto i : S) {
    if (i == '0') {
      ones.push_back(0);
    } else {
      ones.back()++;
    }
  }
  cout << f(int(ones.size()) - 1, 0, 0) << "\n";
  return 0;
}