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: Runtime Error)

using System;
using System.Linq;
using System.Collections.Generic;

namespace AtCoder.Beginner066
{
    public class ABC066D_11
    {
        static int modVal = 1000000007;

        public static void Main(string[] args)
        {

            // データ数の入力
            int N = int.Parse(Console.ReadLine());

            // データの入力
            int[] nums = Console.ReadLine().Split(' ').Select(int.Parse).ToArray();

            // 重複データの確認
            int[] used = new int[nums.Length];
            int duplicateLeft = 0;
            int duplicateRight = 0;
            for(int i = 0; i <= N; i++)
            {
                if(used[nums[i]] != 0)
                {
                    duplicateLeft = used[nums[i]] - 1;
                    duplicateRight = i;
                    break;
                }
                used[nums[i]] = i + 1;
            }

            // パターン数の出力
            for(int i = 1; i <= N + 1; i++)
            {
                long patternNum = GetCombinationNum(N + 1, i);
                patternNum -= GetCombinationNum(duplicateLeft - 1 + N - duplicateRight + 1, i - 1);
                patternNum = (patternNum + modVal) % modVal;

                patternNum = MathMod.Combination(N + 1, i);
                patternNum -= MathMod.Combination(duplicateLeft - 1 + N - duplicateRight + 1, i - 1);
                patternNum = (patternNum + modVal) % modVal;

                Console.WriteLine(patternNum);
            }

        }

        static long GetCombinationNum(int n, int r)
        {
            double patternNum = 1;
            for(int i = n; i > n - r; i--)
            {
                patternNum = patternNum * i;
            }

            for(int i = 2; i <= r; i++)
            {
                patternNum = patternNum / i;
            }

            return (long)(patternNum % modVal);
        }

        public class MathMod
        {

            // 階乗計算に使う素数(階乗計算の上限数) 10^9+7
            static int modVal = 1000000007;

            // 階乗計算を行う数
            private static int FactorialMax = 10005;
        

            // 階乗の剰余
            private static int[] factorialMod = new int[FactorialMax + 1];
        
            // 階乗の逆元
            private static int[] factorialInverse = new int[FactorialMax + 1];


            static MathMod()
            {

                // 階乗の剰余・逆元を計算
                factorialMod[0] = 1;
                factorialInverse[0] = 1;
                for(long i = 1; i <= FactorialMax; i++)
                {
                    factorialMod[i] = (int)((factorialMod[i - 1] * i) % modVal);
                    factorialInverse[i] = Pow(factorialMod[i], modVal - 2);
                }
            
            }

            /// <summary>累乗を求める</summary>
            public static int Pow(int x, int y)
            {
                long remaining = y;
                long p = x;

                long ret = 1;

                while( remaining > 0 )
                {
                    if( remaining % 2 == 0 )
                    {
                        p = (p * p) % modVal;
                        remaining /= 2;
                    }
                    else
                    {
                        ret = (ret * p) % modVal;
                        remaining -= 1;
                    }
                }

                return (int)ret;
            }


            /// <summary>階乗を求める</summary>
            public static double Factorial(int n)
            {
                if(n < 0 || n > FactorialMax)
                {
                    throw new ArgumentOutOfRangeException();
                }
            
                return factorialMod[n];
            }
        

            /// <summary>組み合わせ数を求める</summary>
            /// <param name="n">全体集合の要素数</param>
            /// <param name="r">取り出す要素数</param>
            /// <returns>組み合わせ数</returns>
            public static int Combination(int n, int r)
            {
                if(n < 0 || n > FactorialMax)
                {
                    throw new ArgumentOutOfRangeException();
                }
                if(r < 0 || r > FactorialMax)
                {
                    throw new ArgumentOutOfRangeException();
                }

                if(n < r)
                {
                    return 0;
                }

                return (int)((long)factorialMod[n] * factorialInverse[r] % modVal * factorialInverse[n - r] % modVal);
            }
        
        }

    }

}