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 PHP7 (7.0.15) (judge verdict: Accepted)

<?php
// 答えだすよう
function answer($v) {
	echo "{$v}\n";
	exit(0);
}

//============================================================
// 素数を法とした値を出す用
//============================================================
const mod = 1000000007;

function MintInit(int $a) {
	return(($a % mod + mod) % mod);
}

function MintAdd(int $a, int $b) {
	return(MintInit($a + $b));
}

function MintSub(int $a, int $b) {
	return(MintInit($a - $b));
}

function MintMul(int $a, int $b) {
	return(MintInit($a * $b));
}

function MintDiv(int $a, int $b) {
	return(MintMul($a, modPow($b, mod - 2, mod)));
}
//============================================================

// 繰り返し二乗法
function modPow(int $a, int $n, int $p) {
	if($n == 1) {
		return($a % $p);
	}
	if($n % 2 == 1) {
		return(($a * modPow($a, $n - 1, $p)) % $p);
	}
	
	$t = modPow($a, $n / 2, $p);
	return(($t * $t) % $p);
}

//============================================================
// コンビネーション用
class combination {
	public $fact = [];
	public $ifact = [];
	public function __construct(int $n) {
		$this->fact[0] = 1;
		for($i = 1; $i <= $n; $i++) {
			$this->fact[$i] = MintMul($this->fact[$i - 1], $i);
		}
		// 逆数を取る
		$this->ifact[$n] = MintDiv(1, $this->fact[$n]);
		// 逆数をとったのでかけていけば良い
		for($i = $n; $i >= 1; $i--) {
			$this->ifact[$i - 1] = MintMul($this->ifact[$i], $i);
		}
	}
	public function c(int $n, int $k) {
		if($k < 0 || $k > $n) {
			return(0);
		}
		return(MintMul($this->fact[$n], MintMul($this->ifact[$k], $this->ifact[$n - $k])));
	}
}
//============================================================


// 入力
fscanf(STDIN, "%d %d", $n, $k);

// kは0の個数の最大になるので丸める
$k = min($k, $n - 1);

// クラス生成
$con = new combination($n);

$ans = 0;

// 0の個数(最小移動回数)を固定してパターン数を求める
for($i = 0; $i <= $k; $i++) {
	// 0の場所のパターン数 * 0以外の組み合わせ
	$ans = MintAdd($ans, MintMul($con->c($n, $i), $con->c(($n - $i - 1) + $i, $i)));
}

answer($ans);