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>

#include <ext/pb_ds/assoc_container.hpp> // Common file
#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
using namespace std;

using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vb = vector<bool>;
using vd = vector<double>;
using vs = vector<string>;

typedef tree<
ll,
null_type,
less<ll>,
rb_tree_tag,
tree_order_statistics_node_update>
ordered_set;
// ordered_set X;
// *X.find_by_order(i) = ith largest element (counting from zero)
// X.order_of_key(k) = number of items in X < k (strict lt)

ll MOD = 998244353;

ll pow(ll a, ll x, ll MOD) {
    if (x == 0) return 1;
    ll halfpow = pow(a, x/2, MOD);
    ll ans = (halfpow * halfpow) % MOD;
    if (x % 2 == 1) ans = (ans * a) % MOD;
    return ans;
}

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int N;
    cin >> N;
    
    vi D(N);
    for(int i=0;i<N;i++) cin >> D[i];

    if (D[0] != 0) {
        cout << 0 << endl;
        return 0;
    }

    vll cnts(N+1, 0);
    for (int d : D) {
        cnts[d]++;
    }

    if (cnts[0] != 1) {
        cout << 0 << endl;
        return 0;
    }


    ll ans = 1;
    for (int i=0;i<N;i++) {
        //cout << cnts[i] << endl;
        ans *= pow(cnts[i], cnts[i+1], MOD);
        ans %= MOD;
    }

    cout << ans << endl;

    return 0;
}