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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std ;
#define rep( i, s, t ) for( register int i = s; i <= t; ++ i )
#define re register
#define int long long
int gi() {
	char cc = getchar() ; int cn = 0, flus = 1 ;
	while( cc < '0' || cc > '9' ) {  if( cc == '-' ) flus = - flus ; cc = getchar() ; }
	while( cc >= '0' && cc <= '9' )  cn = cn * 10 + cc - '0', cc = getchar() ;
	return cn * flus ;
}
const int N = 2e5 + 5 ; 
const int P = 1e9 + 7 ; 
int n, m, fac[N], inv[N], Ans ;  
int fpow( int x, int k ) {
	int ans = 1, base = x ; 
	while( k ) {
		if( k & 1 ) ans = ans * base % P ;
		base = base * base % P, k >>= 1 ;
	} return ans ; 
}
int C( int x, int y ) {
	if( y > x ) return 0 ;
	return fac[x] * inv[y] % P * inv[x - y] % P ; 
}
signed main()
{
	n = gi(), m = gi(), inv[0] = fac[0] = 1 ;
	rep( i, 1, n ) fac[i] = fac[i - 1] * i % P, inv[i] = fpow( fac[i], P - 2 ) ;
	Ans = 1 ;
	for( re int i = 1; i <= min( n - 1, m ); ++ i ) {
		int ans = C( n, i ) ;
		int t = n - i, uu = C( n - 1, n - t )  ; 
		Ans = ( Ans + ans * uu % P ) % P ;
	} cout << Ans << endl ;
	return 0 ;
}