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 <algorithm>
#include <cstring>
#include <vector>

#define rep(i, a, b) for ( int i = (a); i < (b); i++ )
#define per(i, a, b) for ( int i = (b)-1; i >= (a); i--)
#define pb push_back
#define bg begin()
#define en end()
#define DEBUG_INT(var) printf("the value of "#var" is %d\n", var)
#define PRINT_ARRAY(name, i) printf("the value of "#name"[%d] is %d\n", (i), name[(i)])
#define DEBUG_ARRAY(name, begin, end) REP(i, begin, end) PRINT_ARRAY(name, i)

using namespace std;

static const int MOD = 998244353;

typedef long long ll;

// 0
// 1 1
// 2 2 2

//   1   1通り
// 4   7 1通り
// 3 5 6 2*2*2 = 8通り
//   2   3  3通り
// 8 9 10

int N;
ll ans;
int D[100005];
ll C[100005];

int main(void) {
    scanf("%d", &N);
    rep(i, 0, N) scanf("%d", &D[i]);

    if (D[0] != 0) {
      printf("0\n");
      return 0;
    }

    rep(i, 1, N) {
      if (D[i] == 0) {
        printf("0\n");
      }
    }

    rep(i, 0, N) C[D[i]] += 1LL;
    
    int depth = -1;
    rep(i, 0, N) {
      if ( D[i] > depth) depth = D[i];
    } 

    ans = C[0];
    rep(i, 1, depth+1) {
      ll tmp = 1;
      rep(j, 0, C[i]) {
        tmp *= C[i-1];
        tmp %= MOD;
      }
      ans *= tmp;
      ans %= MOD;
    }


    printf("%lld\n", ans);

	return 0;
}