Problem:
Score : 200 points

Problem Statement

You are given a string S of length N.
Among its subsequences, count the ones such that all characters are different, modulo 10^9+7. Two subsequences are considered different if their characters come from different positions in the string, even if they are the same as strings.

Here, a subsequence of a string is a concatenation of one or more characters from the string without changing the order.

Constraints

1 \leq N \leq 100000

S consists of lowercase English letters.

|S|=N

Input

Input is given from Standard Input in the following format:

N
S

Output

Print the number of the subsequences such that all characters are different, modulo 10^9+7.

Sample Input 1

4
abcd

Sample Output 1

15

Since all characters in S itself are different, all its subsequences satisfy the condition.

Sample Input 2

3
baa

Sample Output 2

5

The answer is five: b, two occurrences of a, two occurrences of ba. Note that we do not count baa, since it contains two as.

Sample Input 3

5
abcab

Sample Output 3

17

Solution in C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

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

namespace atcorder_GrandContest_031
{
    class Program
    {

        static void Main(string[] args)
        {

            int cnt = int.Parse(Console.ReadLine());
            var data = Console.ReadLine().ToArray();

            List<int> charList = new List<int>();
            int[] cntArr = new int[128];

            int charCnt = 0;

            for (int i = 0; i < cnt; i++)
            {
                int aVal = (int)data[i];
                if (0 == cntArr[aVal])
                {
                    charList.Add(aVal);
                    charCnt++;
                }
                cntArr[aVal]++;
            }

            //文字の種類分選択するパターンの数
            int allCnt = 0;
            //その他のパターンの数
            int otherCnt = 0;

            //文字が複数種類ある場合
            if (1 < charCnt)
            {
                //文字の種類分選択するパターンの計算
                allCnt = 1;
                for (int i = 0; i < charCnt; i++)
                {
                    allCnt *= cntArr[charList[i]];
                }

                //その他のパターンの数の計算
                //各文字が一つずつしかない場合
                if (1 == allCnt)
                {
                    for (int i = charCnt - 1; 2 <= i; i--)
                    {
                        otherCnt += GetCombination(charCnt, i);
                    }
                }
                else
                {
                  
                    for (int i = charCnt - 1; 2 <= i; i--)
                    {
                        
                        int minusVal = 0;

                        for (int j = 0; j < charCnt; j++)
                        {
                            if (1 < cntArr[charList[j]])
                            {
                                //GetCombinationの第二引数の-1は「cntArr[charList[j]]」の分の減算
                                otherCnt += cntArr[charList[j]] * GetCombination(charCnt - minusVal, i - 1);
                                minusVal += cntArr[charList[j]];
                            }
                        }
                    }

                }

            }

            //合算
            int ret = 0;
            //cntは一つずつ選択するパターン
            ret = allCnt + otherCnt + cnt;

            Console.WriteLine(ret);

        }

        //組み合わせ取得
        public static int GetCombination(int charCnt, int selectCnt)
        {
            //分子
            int molecule = 1;
            //分母
            int denominator = 1;

            for (int i = 0; i < selectCnt; i++)
            {
                molecule *= charCnt - i;
                denominator *= selectCnt - i;
            }
            return molecule / denominator;
        }
    }
}