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;
#define ll long long
#define REP(i,m,n) for(int i=(int)(m); i<(int)(n); i++)
#define rep(i,n) REP(i,0,n)
#define RREP(i,m,n) for(int i=(int)(m); i>=(int)(n); i--)
#define rrep(i,n) RREP(i,n-1,0)
#define all(v) v.begin(), v.end()
const int inf = 1e9+7;
const ll longinf = 1LL<<60;
const ll mod = 998244353;

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  string S;
  ll K;
  cin >> S >> K;
  int N = S.size();
  int cnt = 0;
  vector<int> C;
  rep(i, N) {
    if(S[i] == '1') cnt++;
    else {
      C.push_back(cnt);
      cnt = 0;
    }
  }
  C.push_back(cnt);
  reverse(all(C));
  int n = C.size();
  int m = accumulate(all(C), 0);
  vector<vector<vector<ll>>> dp(n+1, vector<vector<ll>>(m+1, vector<ll>(m+1)));
  dp[0][0][0] = 1;
  rep(i, n) { // i番目の区間
    rep(j, m+1) { // 取り出した回数
      if(j > K) continue;
      rep(k, m+1) { // 挿入した回数
        int rem = j-k;
        if(rem < 0) continue;
        rep(l, rem+1) { // ここで挿入する回数
          (dp[i+1][j][k+l] += dp[i][j][k]) %= mod;
        }
        rep(l, C[i]+1) { // ここで取り出す回数
          if(j+l > m) continue;
          if(l == 0) continue;
          (dp[i+1][j+l][k] += dp[i][j][k]) %= mod;
        }
      }
    }
  }
  ll ans = 0;
  rep(i, min(K+1, (ll)m+1)) (ans += dp[n][i][i]) %= mod;
  cout << ans << endl;
  return 0;
}