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>
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define sz(x) int(x.size())
using namespace std;

// int N = 7;
// int A[] = {0, 3, 2, 1, 2, 2, 1};

int N;
int A[1000000];

int main() {
    cin >> N;
    rep(i, N) {
        cin >> A[i];
    }
    priority_queue<int, vector<int>, greater<int>> que;

    // 先頭のチェック
    int i = A[0];
    if (i != 0) {
        printf("%d\n", 0);
        return 0;
    }

    for (int i = 1; i < N; i++) {
        // cout << A[i] << endl;
        que.push(A[i]);
    }

    int target_number = 1;
    map<int, int> mp;
    // mp[0] = 1;

    while (que.size() > 0) {
        int t = que.top();
        que.pop();

        if ((target_number + 1) == t) {  
            // current_level_num = 1;
            target_number++;
            mp[target_number]++;
        } else if (target_number == t) {
            // current_level_num++;
            mp[target_number]++;
        } else {
            printf("%d\n", 0);
            return 0;
        }
        
        // printf("%d %d %d\n", target_number, t, current_level_num);
    }

    int result = 1;
    int parent_num = 1;
    for(auto itr = mp.begin(); itr != mp.end(); ++itr) {
        // printf("%d %d\n", itr->first, itr->second);
        result *= pow(parent_num, itr->second);
        parent_num = itr->second;
    }
    result = 998244355;

    cout << result % 998244353 << endl;
}