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<bits/stdc++.h>

using namespace std;
typedef long long ll;
#define rep(i, n) for (ll i = 0; i < (ll)(n); ++i)
#define erep(i, n) for (ll i = 0; i <= (ll)(n); ++i)
#define FOR(i,a,b) for (ll i = (a); i < (ll)(b); ++i)
#define EFOR(i,a,b) for (ll i = (a); i <= (ll)(b); ++i)
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } }

const ll sub = 998244353;

int pow_kai(ll a, ll n){//aのn乗を計算します。
    int x = 1;
    while(n > 0){//全てのbitが捨てられるまで。
        if(n&1){//1番右のbitが1のとき。
            x = x*a%sub;
        }
        a = a*a%sub;
        n >>= 1;//bit全体を右に1つシフトして一番右を捨てる。
    }
    return x;
}

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;
    }

    vector<int> level(n,0);
    rep(i, n) {
        level[d[i]]++;
    }

    ll ans = 1;
    for(int i=1; i<n; i++) {
        if(level[i]!=0) {
            if(level[i-1] == 0) {
                cout << 0 << endl;
                return 0;
            }
            ans = ans*pow_kai(level[i-1], level[i])%sub;
        }
    }

    cout << ans%sub << endl;

    return 0;

}