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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

#define MOD 1000000007
#define AMAX (2<<17)

int N,K;
long F[AMAX];
long I[AMAX];

long powmod( long, long );

/***********************************************************
    階乗と逆数のテーブル生成
***********************************************************/
void maketable()
{
    // 階乗を計算し記録
    F[0] = 1;
    for( int i=1; i<=N+1; i++ )
        F[i] = (long long) F[i-1] * i % MOD;

    // 階乗の逆数を計算し記録
    I[N] = powmod( F[N], MOD -2 );
    for( int i=N-1; i>=0; i-- )
        I[i] = (long long) I[i+1] * (i+1) % MOD;
}

/***********************************************************
    べき乗＋剰余
***********************************************************/
long powmod( long a, long b )
{
    long long ret = 1;

    while( b ) {
        if( b & 1 ) ret = ret * a % MOD;
        b >>= 1;
        a = (long long)a * a % MOD;
    }

    return( ret );
}

/***********************************************************
    組合せ＋剰余
***********************************************************/
long combination( long n, long r )
{
    long long ret;

    ret = F[n];
    ret = (ret * I[n-r]) % MOD;
    ret = (ret * I[r]) % MOD;

    return( ret );
}

/***********************************************************

***********************************************************/
int main( int argc, char *argv[] )
{
    // 問題を読み取る
    scanf( "%d%d", &N, &K );

    // 階乗と逆数のテーブルを生成
    maketable();

    // ΣnCr * n-1Cr を計算
    long long ans = 0;
    for(long i=0;i<N&&i<=K;i++) {
//        ans += F[N] * F[N-1] % MOD * I[N-i] % MOD * I[i] % MOD * I[N-1-i] % MOD * I[i] % MOD;
        ans += (long long) combination( N, i ) * combination( N-1, i ) % MOD;
    }

//    while( ans < 0 ) ans += MOD;
    ans %= MOD;

    printf( "%ld\n", (long)ans );
}