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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>

int cmp(const int *a,const int *b){
  if(*a>*b){
    return 1;
  }
  else if(*a<*b)
  return -1;
  else return 0;
}

int main(void){
  int n;

  scanf("%d",&n);

  int d[n];
  long long kosu[n];
  long long ans=1;
  int cnt;
  int i,j,k;

  for(i=0;i<n;i++){
    scanf("%d",&d[i]);
    if(d[i]==0&&i!=0){
      printf("0\n");
      return 0;
    }
  }
  if(d[0]!=0){
    printf("0");
    return 0;
  }




  qsort(d,n,sizeof(int),(int (*)(const void *,const void *))cmp);
/*for(i=0;i<n;i++){
  printf("%d",d[i]);
}*/
for(i=0;i<n-1;i++){
if(d[i+1]-d[i]>1){
  printf("%d",0);
  return 0;
}
}
k=1;
kosu[0]=1;
  for(i=1;i<n;i=i+cnt){
    cnt=0;
    for(j=i;j<n;j++){
      if(d[j]==d[j+1]){
        cnt++;
        ans=ans*kosu[k-1]%998244353;
      }
      else{
        ans=ans*kosu[k-1]%998244353;
        kosu[k]=cnt+1;
        k++;
        cnt++;
        break;
      }
    }
  }
/*for(i=0;i<k;i++){
  printf("kosu[%d]=%d\n",i,kosu[i]);
}*/

/* for(i=1;i<k;i++){

    ans=ans*pow(kosu[i-1],kosu[i]);

    ans=ans%998244353;

  }*/
  ans=ans%998244353;
  printf("%lld",ans);



return 0;
}