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: Wrong Answer)

#include <bits/stdc++.h>
#define LL long long
using namespace std;
const int P = 1e9 + 7;
inline int qpow(int x,int y){
	static int r; r=1; while (y){ if (y&1) r=(LL)r*x%P;y>>=1,x=(LL)x*x%P; } return r;
}
inline int Inv(int x){
	return qpow(x,P-2);
}
int fac[800050],nfac[800050],inv[800050];
inline int C(int n,int m){
	return (m>n||n<0||m<0)?(0):((LL)fac[n]*nfac[m]%P*nfac[n-m]%P); 
}
int n,k,ans;
int main(){
	cin>>n>>k;
	if (k >= n){ cout << C(2*n-1,n-1) << endl; return 0; }
	int a = k,b = n-k,i;
	fac[0] = nfac[0] = inv[0] = fac[1] = nfac[1] = inv[1] = 1;
	for (i = 2; i <= 4 * n; ++i){
		fac[i] = (LL)fac[i-1] * i % P;
		inv[i] = (LL)(P-P/i) * inv[P%i] % P;
		nfac[i] = (LL)nfac[i-1] * inv[i] % P;
	}
	for (i = 0; i <= n; ++i){
		a = n-i; if (!a || a<b) continue;
		ans += (LL)C(n-1,a-1) * C(n,i) % P; ans%=P;
	}
	cout << ans << endl;
    return 0;
}