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 <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <cstdlib>
#include <cmath>
using namespace std;
int main(){
  long long mod = 1000000007;
  int kazu = 0;
  int idou = 0;
  cin >> kazu >> idou;
  vector<long long> jyou(400001);
  jyou[1] = 1;
  for(int i = 2;i <= 400000;i++){
    jyou[i] = (jyou[i - 1] * i) % mod;
  }
  //cout << 1 << endl;
  long long out = 0;
  if(idou + 1 < kazu){
    out = (jyou[(kazu - idou) + (kazu - 1)]  / jyou[kazu - idou] / jyou[kazu - 1]) % mod;
  }
  //cout << 2 << endl;
  long long kotae = 0;
  //cout << jyou[kazu * 2 - 1] << endl;
  //cout << jyou[kazu] << endl;
  //cout << jyou[kazu - 1] << endl;
  long long kamo = 0;
  long long kamoo = 0;
  kotae = (((jyou[kazu * 2 - 1] / jyou[kazu] ) % mod) / jyou[kazu - 1]) % mod;
  //cout << 3 << endl;
  //cout << kotae << endl;
  cout << kotae - out << endl;
  return 0;
}