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: Wrong Answer)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <ctype.h>
#include <math.h>
#define MAX(X, Y) (((X) > (Y))? (X) : (Y))
#define MIN(X, Y) (((X) < (Y))? (X) : (Y))
int compare_int(const void *a, const void *b){
  return *(long int*)b - *(long int*)a; // 大きい順
}
int compare_int_small(const void *a, const void *b){
  return *(long int*)a - *(long int*)b; // 小さい順
}
// 公約数
long long int gcd(long long int a, long long int b){
  if(b == 0){
    return a;
  }else{
    return gcd(b, a % b);
  }
}
// 最小公倍数 = a * b / 最大公約数
// 公倍数
long long int lcm(long long int a, long long int b){
  return a * b / gcd(a, b);
}

#define TRUE 1
#define FALSE 0
#define MAX_NUM 2502
typedef struct pp{
  char c;
  int h;
  int w;
}P;

int top = 0;
P *Stack = NULL;
P *tmp = NULL;


void initialize(){
  Stack = (P *)calloc(MAX_NUM, sizeof(P));
  top = 0;
}

void finalize(){
  if(Stack != NULL){
    free(Stack);
    Stack = NULL;
  }
}

int isEmpty(){
  if(top < 0){
    return TRUE;
  }else{
    return FALSE;
  }
}

int isFull(){
  if(top >= MAX_NUM - 1){
    return TRUE;
  }else{
    return FALSE;
  }
}

int push(P x){
  fprintf(stderr, "pushするのは%c (h, w) = (%d, %d)\n", x.c, x.h, x.w);
  //printf("push x = %d\n",x);
  if(isFull() != FALSE){
    //    fprintf(stderr, "pushしたのは%d top = %d\n", Stack[top], top);
    printf("Error! Overfllow\n");
    return 0;
  }else{
    top++;
    if(Stack != NULL){
      Stack[top] = x;
    }
    return 0;
  }
}

P pop(){
  if(isEmpty() != FALSE){
    printf("Error! Underfllow\n");
    return *tmp;
  }else{
    top--;
    // return では取りだす値を返す
    //    fprintf(stderr, "popしたのは%d top = %d\n", Stack[top + 1], top);
    return Stack[top+1];
  }
}

// min
int main(void){
  int n, d;
  long ans = 1;
  int *A;
  int max = 0;

  scanf("%d", &n);
  //  printf("n = %d\n", n);
  
  A = (int *)calloc(n, sizeof(int));
  if(A == NULL){
    fprintf(stderr, "メモリが確保できません.\n");
    exit(EXIT_FAILURE);
  } 

  //  printf("n = %d\n", n);
  
  for(int i = 0; i < n; i++){
    scanf("%d", &d);
    A[d]++;
    //    printf("d = %d n = %d\n", d, n);
    if(max < d){
      max = d;
    }
  }

  if(A[0] == 0){
    printf("0\n");
    return 0;
  }
  if(n >= 1){
    for(int i = 1; i <= max; i++){
      for(int j = 0; j < A[i]; j++){
	if(ans > 998244353){
	  ans %= 998244353;
	}
	if(A[i - 1] > 998244353){
	  A[i - 1] %= 998244353;
	}
	ans *= A[i - 1];
	if(ans > 998244353){
	  ans %= 998244353;
	}
      }
  
    }
  }


  printf("%ld\n", ans);
  if(A != NULL){
    free(A);
    A = NULL;
  }
  
  return 0;

}