Problem:
Score : 300 points

Problem Statement

Given is an integer sequence D_1,...,D_N of N elements. Find the number, modulo 998244353, of trees with N vertices numbered 1 to N that satisfy the following condition:

For every integer i from 1 to N, the distance between Vertex 1 and Vertex i is D_i.

Notes

A tree of N vertices is a connected undirected graph with N vertices and N-1 edges, and the distance between two vertices are the number of edges in the shortest path between them.

Two trees are considered different if and only if there are two vertices x and y such that there is an edge between x and y in one of those trees and not in the other.

Constraints

1 \leq N \leq 10^5

0 \leq D_i \leq N-1

Input

Input is given from Standard Input in the following format:

N
D_1 D_2 ... D_N

Output

Print the answer.

Sample Input 1

4
0 1 1 2

Sample Output 1

2

For example, a tree with edges (1,2), (1,3), and (2,4) satisfies the condition.

Sample Input 2

4
1 1 1 1

Sample Output 2

0

Sample Input 3

7
0 3 2 1 2 2 1

Sample Output 3

24

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <cstring>
#include <map>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <set>
#include <vector>
#include <queue>
#include <list>
#include <numeric>
#include <stdio.h>
#include <string>
#include <cstdlib>
#include <math.h>
#include <stack>
#include <climits>
#include <bitset>
#include <utility>


using namespace std;
typedef long long ll;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
// const ll INF = 1LL <<60;
const int INF = 10000;

//最大公約数
ll gcd(ll x, ll y) {
  ll tmp = 0;
  if (x < y){
    tmp=x;
    x=y;
    y=tmp;
  }
  while (y > 0) {
    ll r = x % y;
    x = y;
    y = r;
  }
  return x;
}

//最大公倍数
ll lcm(ll x,ll y){
  return x/gcd(x,y)*y;
}

const int MAX = 510000;
const int MOD = 1000000007;

long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; 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;
    }
}

// 二項係数計算
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}


//階乗
ll kaijo(ll k){
    ll sum = 1;
    for (ll i = 1; i <= k; ++i)
    {
        sum *= i;
        sum%=1000000000+7;
    }
    return sum;
}

//for(int i = ; i < ; i++){}

ll lmax(ll s,ll t){
  if(s>t){
    return s;
  }
  else{
    return t;
  }
}

ll lmin(ll s,ll t){
  if(s<t){
    return s;
  }
  else{
    return t;
  }
}
// ここから開始
int main(){
  ll n;
  // n = 1e5;
  cin>>n;
  ll cnt[n];
  for(int i = 0; i < n; i++) cnt[i]=0;
  for(int i = 0; i < n; i++){
    int x;
    cin>>x;
    cnt[x]++;
  }
  
  // cnt[0]=1;
  // cnt[1]=1e5-1;
  // cnt[2]=1e5-1;
  ll ans = 1;
  ll mod = 998244353;
  bool flg = true;
  if(cnt[0]!=1) ans=0;
  for(int i = 1; i < n; i++){
    if(cnt[i]==0){
      flg = false;
    }
    else{
      if(!flg){
        ans*=0;
      }
      else{
        ll x=1;
        for(int j = 0; j < cnt[i]; j++){
          x*=cnt[i-1];
          x%=mod;
        }
        // cout<<x<<endl;
        x%=mod;
        ans*=x;
        ans%=mod;
        // cout<<ans<<endl;
      }
    }
    ans%=mod;
    // cout<<ans<<endl;
  }
  cout<<ans<<endl;
  return 0;
}