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: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> l_l;
typedef pair<int, int> i_i;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))

const ll mod = 998244353;

ll safePow(ll x, ll y){
    ll result = 1;
    rep(i, y){
        result = (result * x) % mod;
    }
    return result;
}

int main() {
    int n;
    cin >> n;
    vector<int> d(n);
    rep(i, n) cin >> d[i];
    if(d[0] != 0){
        cout << 0 << endl;
        return 0;
    }
    sort(d.begin(), d.end());
    if(d[1] != 1){
        cout << 0 << endl;
        return 0;
    }

    ll cnt = 0, cur_d = 0, prev_cnt = 1;
    ll ans = 1;
    rep(i, n){
        if(d[i] == cur_d){
            cnt++;
        } else {
            if(d[i] != cur_d + 1){
                ans = 0;
                break;
            }
            ans = (ans * safePow(prev_cnt, cnt)) % mod;
            prev_cnt = cnt;
            cnt = 1;
            cur_d++;
        }
        if(i == n - 1){
            ans = (ans * safePow(prev_cnt, cnt)) % mod;
        }
    }
    cout << ans << endl;

    
    return 0;
}