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 Kotlin (1.0.0) (judge verdict: Accepted)

fun main(args: Array<String>) {
    val mod = 1000000007L

    val (n, k) = readInputLine().split(" ").map { it.toInt() }

    val factorial = LongArray(n + 1)
    val inverse = LongArray(n + 1)
    val factorialInverse = LongArray(n + 1)
    factorial[0] = 1L
    factorial[1] = 1L
    inverse[0] = 1L
    inverse[1] = 1L
    factorialInverse[0] = 1L
    factorialInverse[1] = 1L
    for (i in 2..n) {
        factorial[i] = (factorial[i - 1] * i.toLong() % mod)
        inverse[i] = (mod - inverse[mod.toInt() % i] * (mod / i.toLong()) % mod)
        factorialInverse[i] = (factorialInverse[i - 1] * inverse[i] % mod)
    }

    fun nCm(n: Int, m: Int): Long {
        return factorial[n] * (factorialInverse[m] * factorialInverse[n - m] % mod) % mod
    }

    fun nHm(n: Int, m: Int): Long {
        return nCm(n + m - 1, m)
    }

    var ans = 0L

    for (i in 0..Math.min(n - 1, k)) {
        ans += nCm(n, i) * nHm(n - i, i) % mod
        ans %= mod
    }

    println(ans)
}

fun readInputLine(): String {
    return readLine()!!
}