Problem:
Score : 900 points

Problem Statement

There are N Snukes lining up in a row.
You are given a string S of length N. The i-th Snuke from the front has two red balls if the i-th character in S is 0; one red ball and one blue ball if the i-th character in S is 1; two blue balls if the i-th character in S is 2.

Takahashi has a sequence that is initially empty. Find the number of the possible sequences he may have after repeating the following procedure 2N times, modulo 998244353:

Each Snuke who has one or more balls simultaneously chooses one of his balls and hand it to the Snuke in front of him, or hand it to Takahashi if he is the first Snuke in the row.

Takahashi receives the ball and put it to the end of his sequence.

Constraints

1 \leq |S| \leq 2000

S consists of 0,1 and 2.

Note that the integer N is not directly given in input; it is given indirectly as the length of the string S.

Input

Input is given from Standard Input in the following format:

S

Output

Print the number of the possible sequences Takahashi may have after repeating the procedure 2N times, modulo 998244353.

Sample Input 1

02

Sample Output 1

3

There are three sequences that Takahashi may have: rrbb, rbrb and rbbr, where r and b stand for red and blue balls, respectively.

Sample Input 2

1210

Sample Output 2

55

Sample Input 3

12001021211100201020

Sample Output 3

543589959

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <algorithm>
#include <bitset>
#include <complex>
#include <deque>
#include <iostream>
#include <istream>
#include <iterator>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
#include <tuple>
#include <cstdint>
#include <iomanip>
using namespace std;
typedef long long ll;
#define rep(i, n) for(ll i = 0; i < (n); i++)
#define revrep(i, n) for(ll i = (n)-1; i >= 0; i--)
#define pb push_back
#define f first
#define s second
void BinarySay(ll x, ll y = 60){rep(i, y) cout << (x>>(y-1-i) & 1); cout << endl;}
const ll INFL = 1LL << 60;//10^18 = 2^60
//ll MOD = 1000000007;
ll MOD = 998244353;
//vector<mint> dp(N, Mint);
//vector<vector<mint>> dp2(N, vector<mint>(N, Mint));
//vector<vector<vector<mint>>> dp3(N, vector<vector<mint>>(N, vector<mint>(N, Mint)));
struct mint{
  ll x;
  mint(ll x=0):x(x % MOD){}
  mint& operator+=(const mint a){
    (x += a.x) %= MOD;
    return *this;
  }
  mint& operator-=(const mint a){
    (x += MOD-a.x) %= MOD;
    return *this;
  }
  mint& operator*=(const mint a){
    (x *= a.x) %= MOD;
    return *this;
  }
  mint& operator%=(const mint a){
    (x %= a.x);
    return *this;
  }
  mint& operator++ (int){
    (x += 1) %= MOD;
    return *this;
  }
  mint& operator-- (int){
    (x += MOD-1) %= MOD;
    return *this;
  }
  mint operator+(const mint a) const{
    mint res(*this);
    return res+=a;
  }
  mint operator-(const mint a) const{
    mint res(*this);
    return res-=a;
  }
  mint operator*(const mint a) const{
    mint res(*this);
    return res*=a;
  }
  mint operator%(const mint a) const{
    mint res(*this);
    return res%=a;
  }
  mint po(ll t) const{
    if(!t) return 1;
    mint a = po(t>>1);
    a *= a;
    if(t&1) a *= *this;
    return a;
  }
  mint inverse() const{
    return po(MOD-2);
  }
  mint& operator/=(const mint a){
    return (*this) *= a.inverse();
  }
  mint operator/(const mint a) const{
    mint res(*this);
    return res/=a;
  }
  bool operator == (const mint a){
    return this->x == a.x;
  }
  bool operator != (const mint a){
    return this->x != a.x;
  }
  void get(){
    cout << x << endl;
  }
  void ge(){
    cout << x << " ";
  }
};const mint Mint = 0;

mint pow_mod(ll x, ll k){
  mint res = 1;
  mint a = x;
  while(k > 0){
    if(k % 2){
      res *= a;
    }
    a *= a;
    k /= 2;
  }
  return res;
}

mint inverse(ll x){return pow_mod(x, MOD-2);}

//二項演算
const int MAXcomb = 200010;
ll fac[MAXcomb], finv[MAXcomb], inv[MAXcomb];
void COMinit(){
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for(int i = 2; i < MAXcomb; i++){
        fac[i] = fac[i-1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD/i) % MOD;
        finv[i] = finv[i-1] * inv[i] % MOD;
    }
}
mint comb(int n, int k){
    if(n < k) return 0;
    if(n < 0 || k < 0) return 0;
    mint res = fac[n];
    res *= finv[k] * finv[n-k];
    return res;
}
mint comb_naive(ll N, ll K){//Kが小さい時
  mint res = 1;
  rep(i, K) res *= (N-i);
  mint k = 1;

  rep(i, K) k *= (i+1);
  res /= k;
  return res;
}
//第二種スターリング数
const ll MAXStir2 = 3010;
vector<vector<mint>> Stir2memo(MAXStir2, vector<mint>(MAXStir2, Mint));
vector<mint> Bellmemo(MAXStir2, Mint);
void Stir2init(){
  Stir2memo[0][0] = 1;
  rep(i, MAXStir2-1)rep(j, i+1)Stir2memo[i+1][j+1] = Stir2memo[i][j] + Stir2memo[i][j+1] * (j+1);
  rep(i, MAXStir2){
    Bellmemo[i] = 0;
    rep(j,i+1) Bellmemo[i] += Stir2memo[i][j];
  }
}
mint Stir2(ll i, ll j){//区別できるi個をjグループに分ける場合の数
  if(i < 0 || j < 0 || i < j) return 0;
  return Stir2memo[i][j];
}
mint Bell(ll x){//区別できるx個をグループ分けする方法全ての場合の数
  if(x < 0) return 0;
  return Bellmemo[x];
}
mint kai_mod(ll K){
  if(K < 0) return 0;
  if(K == 0) return 1;
  return kai_mod(K-1) * K;
}

//約数の列挙O(√n)
vector<ll> divisor(ll n){
    vector<ll> res(0);
    for(ll i = 1; i * i <= n; i++){
        if(n % i == 0){
            res.push_back(i);
            if(i != n/i) res.push_back(n/i);
        }
    }
    sort(res.begin(), res.end());
    return res;
}


//Range Sum Query
const ll N_ = (1 << 18);
ll seg[2*N_-1];
//未確認
void add(ll k, ll x){
  k += N_-1;
  seg[k] += x;
  while(k > 0){
    k = (k-1)/2;
    seg[k] = seg[2*k+1] + seg[2*k+2];
  }
}

ll get_sum(ll x, ll y, ll k = 0, ll left = 0, ll right = N_){
  if(right <= x || y <= left) return 0;
  if(x <= left && right <= y) return seg[k];
  ll mid = (left + right) / 2;
  ll l = get_sum(x, y, 2*k+1, left, mid);
  ll r = get_sum(x, y, 2*k+2, mid, right);
  return l + r;
}

mint dp[2010][2010];
ll N;
ll B;
void solve(){
  dp[0][0] = 1;//i番目まで見て、青をj個使った
  for(ll i = 0; i < 2*N; i++){
    ll can_use_blue = get_sum(0, i+1);
    ll can_use_red = 2 * (i+1) - can_use_blue;
    for(ll j = 0; j <= B; j++){
      if(can_use_blue >= j+1){
        dp[i+1][j+1] += dp[i][j];
      }
      if(can_use_red >= i+1-j ){
        dp[i+1][j] += dp[i][j];
      }
    }
  }
  
  dp[2*N][B].get();
}

int main(){
  string S;
  cin >> S;
  N = S.size();
  rep(i, N){
    if(S[i] == '2') add(i, 2);
    if(S[i] == '1') add(i, 1);
  }
  B = get_sum(0, N_);
  solve();
}