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 Pascal (FPC 2.6.2) (judge verdict: Accepted)

var
	ans,x,y,mo:int64;
	F,I:Array[0..200000]of int64;
	n,k,id:Longint;
function p(a,b:int64):int64;
begin
	if b=0 then p:=1 else if b mod 2=0 then p:=p(a*a mod mo,b div 2) else p:=p(a,b-1)*a mod mo;
end;
begin
	read(n,k);
	F[0]:=1;
	mo:=1000000007;
	for id:=1 to n do F[id]:=F[id-1]*id mod mo;
	I[n]:=p(F[n],mo-2);
	for id:=n-1 downto 0 do I[id]:=I[id+1]*(id+1)mod mo;
	id:=0;
	while(id<n)and(id<=k)do begin
		inc(ans,F[n]*I[n-id]mod mo*I[id]mod mo*F[n-1]mod mo*I[n-1-id]mod mo*I[id]mod mo);
		inc(id);
	end;
	writeln(ans mod mo);
end.