Problem:
Score : 400 points

Problem Statement

There is a knight - the chess piece - at the origin (0, 0) of a two-dimensional grid.

When the knight is at the square (i, j), it can be moved to either (i+1,j+2) or (i+2, j+1).

In how many ways can the knight reach the square (X, Y)?

Find the number of ways modulo 10^9 + 7.

Constraints

1 \leq X \leq 10^6

1 \leq Y \leq 10^6

All values in input are integers.

Input

Input is given from Standard Input in the following format:

X Y

Output

Print the number of ways for the knight to reach (X, Y) from (0, 0), modulo 10^9 + 7.

Sample Input 1

3 3

Sample Output 1

2

There are two ways: (0,0) \to (1,2) \to (3,3) and (0,0) \to (2,1) \to (3,3).

Sample Input 2

2 2

Sample Output 2

0

The knight cannot reach (2,2).

Sample Input 3

999999 999999

Sample Output 3

151840682

Print the number of ways modulo 10^9 + 7.

Solution in Kotlin (1.0.0) (judge verdict: Accepted)

import java.util.*

fun main(args: Array<String>) {
    val result = solve(Scanner(System.`in`))
    if (result.isNotEmpty()) println(result)
}

const val MOD = 1000000007L

fun modMul(a: Long, b: Long): Long = (a * b) % MOD

fun modDiv(a: Long, b: Long): Long = (a * modPow(b, MOD - 2)) % MOD

fun modPow(a: Long, p: Long): Long = when {
    p == 0L -> 1
    p % 2 == 1L -> (a * modPow(a, p - 1)) % MOD
    else -> {
        val half = modPow(a, p / 2)
        (half * half) % MOD
    }
}

fun solve(sc: Scanner): String {
    val X = sc.nextInt()
    val Y = sc.nextInt()

    if ((X + Y) % 3 != 0) {
        return "0"
    }
    val a = ((2 * Y) - X) / 3
    val b = ((2 * X) - Y) / 3
    if (b < 0 || a < 0) {
        return "0"
    }

    var result = 1L
    for (i in (b + 1)..(a + b)) {
        result = modMul(result, i.toLong())
    }
    for (i in (1..a)) {
        result = modDiv(result, i.toLong())
    }

    return result.toString()
}