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 <stdbool.h>
#include <stdint.h>
#include <stdlib.h>
#include <limits.h>
#include <math.h>
int acs(const void *a, const void *b) { return *(int *)a - *(int *)b; }                            /* 1,2,3,4.. */
int des(const void *a, const void *b) { return *(int *)b - *(int *)a; }                            /* 8,7,6,5.. */
int cmp_char(const void *a, const void *b) { return *(char *)a - *(char *)b; }                     /* a,b,c,d.. */
int cmp_str(const void *a, const void *b) { return strcmp(*(const char **)a, *(const char **)b); } /* aaa,aab.. */
#define min(a, b) (a < b ? a : b)
#define max(a, b) (a > b ? a : b)
#define rep(i, l, r) for (int i = l; i < r; i++)
#define MAX 100001
#define MOD 998244353

typedef long long int lli;

int main(void)
{
  int n, d[MAX];
  scanf("%d", &n);

  int kyori[MAX] = {0}; // kyori[i]:頂点1からの距離がiの個数
  int len = 0;
  rep(i, 0, n)
  {
    scanf("%d", &d[i]);
    len = max(len, d[i]);
    kyori[d[i]]++;
  }

  lli ans = 1;
  rep(i, 0, len + 1)
  {
    if (kyori[i] == 0)
    {
      ans = 0;
      break;
    }
    ans = (ans * (lli)pow(kyori[i], kyori[i + 1])) % 998244353;
  }

  printf("%lld\n", ans);
  return 0;
}