Problem:
Score : 500 points

Problem Statement

There is a building with n rooms, numbered 1 to n.

We can move from any room to any other room in the building.

Let us call the following event a move: a person in some room i goes to another room j~ (i \neq j).

Initially, there was one person in each room in the building.

After that, we know that there were exactly k moves happened up to now.

We are interested in the number of people in each of the n rooms now. How many combinations of numbers of people in the n rooms are possible?

Find the count modulo (10^9 + 7).

Constraints

All values in input are integers.

3 \leq n \leq 2 \times 10^5

2 \leq k \leq 10^9

Input

Input is given from Standard Input in the following format:

n k

Output

Print the number of possible combinations of numbers of people in the n rooms now, modulo (10^9 + 7).

Sample Input 1

3 2

Sample Output 1

10

Let c_1, c_2, and c_3 be the number of people in Room 1, 2, and 3 now, respectively. There are 10 possible combination of (c_1, c_2, c_3):

(0, 0, 3)

(0, 1, 2)

(0, 2, 1)

(0, 3, 0)

(1, 0, 2)

(1, 1, 1)

(1, 2, 0)

(2, 0, 1)

(2, 1, 0)

(3, 0, 0)

For example, (c_1, c_2, c_3) will be (0, 1, 2) if the person in Room 1 goes to Room 2 and then one of the persons in Room 2 goes to Room 3.

Sample Input 2

200000 1000000000

Sample Output 2

607923868

Print the count modulo (10^9 + 7).

Sample Input 3

15 6

Sample Output 3

22583772

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

using System;
using System.Numerics;
using System.Linq;
using System.Collections.Generic;
class ProgramE
	{
		static void Main(string[] args)
		{

			//入力
			string[] input = Console.ReadLine().Split(' ');
			long n = long.Parse(input[0]);
			long k = long.Parse(input[1]);

			long MOD = 1000000000 + 7;
			long sum = 0;

			//下準備
			nCk com = new nCk();
			com.fac = new long[n + 1];
			com.inv = new long[n + 1];
			com.finv = new long[n + 1];
			com.mod = MOD;
            com.max = n + 1;

			//メモ作成
			com.com_init();
            
            //最大の部屋の数を考える
            long m = n;
          　if(n > k)
                m = k + 1;

			//nCm×n-mHm = nCm×n-1Cn-m-1の和を出す
			for(long i = 0;i < m;i++)
			{
				sum += com.com(n, i) * com.com(n - 1, n - i - 1) % MOD;
				sum %= MOD;
			}

			//出力
			Console.WriteLine(sum);

		}

	}

	class nCk
	{
		public long mod;
		public long max;
		public long[] fac;
		public long[] inv;
		public long[] finv;
		

		//メモ作成
		public void com_init()
		{
			fac[0] = 1;
			fac[1] = 1;
			finv[0] = 1;
			finv[1] = 1;
			inv[1] = 1;
			for(long i = 2;i < max;i++)
			{
				fac[i] = i * fac[i - 1] % mod;
				inv[i] = mod - inv[mod % i] * (mod / i) % mod;
				finv[i] = finv[i - 1] * inv[i] % mod;
			}
		}

		//二項計算
		public long com(long n,long k)
		{
			if (n < 0 || k < 0)
				return 0;
			if (n < k)
				return 0;
			return fac[n] * (finv[n - k] * finv[k] % mod) % mod;
		}

	}