Problem:
Score : 600 points

Problem Statement

You are given an integer sequence of length n+1, a_1,a_2,...,a_{n+1}, which consists of the n integers 1,...,n.
It is known that each of the n integers 1,...,n appears at least once in this sequence.

For each integer k=1,...,n+1, find the number of the different subsequences (not necessarily contiguous) of the given sequence with length k, modulo 10^9+7.

Notes

If the contents of two subsequences are the same, they are not separately counted even if they originate from different positions in the original sequence.

A subsequence of a sequence a with length k is a sequence obtained by selecting k of the elements of a and arranging them without changing their relative order. For example, the sequences 1,3,5 and 1,2,3 are subsequences of 1,2,3,4,5, while 3,1,2 and 1,10,100 are not.

Constraints

1 \leq n \leq 10^5

1 \leq a_i \leq n

Each of the integers 1,...,n appears in the sequence.

n and a_i are integers.

Input

Input is given from Standard Input in the following format:

n
a_1 a_2 ... a_{n+1}

Output

Print n+1 lines.
The k-th line should contain the number of the different subsequences of the given sequence with length k, modulo 10^9+7.

Sample Input 1

3
1 2 1 3

Sample Output 1

3
5
4
1

There are three subsequences with length 1: 1 and 2 and 3.

There are five subsequences with length 2: 1,1 and 1,2 and 1,3 and 2,1 and 2,3.

There are four subsequences with length 3: 1,1,3 and 1,2,1 and 1,2,3 and 2,1,3.

There is one subsequence with length 4: 1,2,1,3.

Sample Input 2

1
1 1

Sample Output 2

1
1

There is one subsequence with length 1: 1.

There is one subsequence with length 2: 1,1.

Sample Input 3

32
29 19 7 10 26 32 27 4 11 20 2 8 16 23 5 14 6 12 17 22 18 30 28 24 15 1 25 3 13 21 19 31 9

Sample Output 3

32
525
5453
40919
237336
1107568
4272048
13884156
38567100
92561040
193536720
354817320
573166440
818809200
37158313
166803103
166803103
37158313
818809200
573166440
354817320
193536720
92561040
38567100
13884156
4272048
1107568
237336
40920
5456
528
33
1

Be sure to print the numbers modulo 10^9+7.

Solution in C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.Linq;
using System.Text;
using static System.Math;

class Program
{
    static int n;
    static int[] a;
    static long mod = (long)1e9 + 7;

    static void Main()
    {
        /* 入力を受け取る */
        n = int.Parse(Console.ReadLine());
        a = Console.ReadLine().Split().Select(int.Parse).ToArray();

        /* 重複について
         *   l個 重複左 (n - l - r)個 重複右 r個
         *  なるl,rを調べる */
        bool[] check = new bool[n + 1]; // check[i] : iがすでに登場したか
        int d = 0; // 重複する数を格納
        for (int i = 0; i <= n; i++)
        {
            if (check[a[i]]) // すでに登場していれば重複
            {
                d = a[i];
                break;
            }
            check[a[i]] = true; // 登場済みに
        }
        int l = 0;
        int r = 0;
        for (; l <= n; l++) // lを決定
            if (a[l] == d)
                break;
        for (; r <= n; r++) // rを決定
            if (a[n - r] == d)
                break;

        /* Combination計算の前処理 */
        Comb c = new Comb(n, mod);

        /* 各kについて答えを計算 */
        for (int k = 1; k <= n + 1; k++)
        {
            long ans = 0;
            ans += c.Com(n - 1, k);
            ans += 2 * c.Com(n - 1, k - 1) % mod;
            ans %= mod;
            ans -= c.Com(l + r, k - 1);
            ans += mod;
            ans %= mod;
            ans += c.Com(n - 1, k - 2);
            ans %= mod;
            Console.WriteLine(ans);
        }
    }
}

class Comb // 組み合わせ計算用のクラス
{
    private long size;
    private long[] inv;
    private long[] kai;
    private long[] invkai;
    private long mod;
    public Comb(long size, long mod)
    {
        this.size = size;
        this.mod = mod;
        inv = new long[size + 1];
        kai = new long[size + 1];
        invkai = new long[size + 1];
        inv[1] = 1;
        for (int i = 2; i <= size; i++)
            inv[i] = mod - (mod / i) * inv[mod % i] % mod;
        kai[0] = 1;
        invkai[0] = 1;
        for (int i = 1; i <= size; i++)
        {
            kai[i] = kai[i - 1] * i % mod;
            invkai[i] = invkai[i - 1] * inv[i] % mod;
        }
    }
    public long Com(long a, long b)
    {
        if (a < b || b < 0)
            return 0;
        return (invkai[b] * invkai[a - b]) % mod * kai[a] % mod;
    }
}