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 N = 303;
const int MOD = 998244353;

int n, k;
char s[N];

vector<int> ones;

int dp[N][N][N];
int f(int x, int use, int sum) {
  if (use > k) return 0;
  if (x < 0) return sum == 0;
  if (~dp[x][use][sum]) return dp[x][use][sum];

  int ret = f(x-1, use, sum);
  for (int i = 1; i <= ones[x]; ++i) {
    ret = (ret + f(x-1, use + i, sum + i)) % MOD;
  }
  for (int i = 1; i <= sum; ++i) {
    ret = (ret + f(x-1, use, sum - i)) % MOD;
  }

  return dp[x][use][sum] = ret;
}

int solve() {
  scanf("%s %d", s, &k);
  
  int n = strlen(s);
  ones.push_back(0);
  for (int i = 0; i < n; ++i) {
    if (s[i] == '0') {
      ones.push_back(0);
    } else {
      ++ones.back();
    }
  }

  // for (int o: ones) cerr << o << " "; cerr << endl;

  memset(dp, -1, sizeof dp);
  printf("%d\n", f((int) ones.size() - 1, 0, 0));
  return 0;
}

int main() {
  int t = 1;
  // scanf("%d", &t);
  while (t--) solve();
  return 0;
}