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: Wrong Answer)

fun main(args: Array<String>) {
    abc145d()
}

private fun mod(x: Long) = x % (1e9 + 7).toLong()

// a,b(初期0,0)を1,2or2,1ずつ増やしてx,yと一致させると考える
// 行けるところはx+yが3の倍数
fun abc145d() {
    val (x, y) = readLine()!!.split(' ').map { it.toLong() }

    if (x * 2 < y || x > y * 2) return println(0)

    if ((x + y) % 3 != 0L) return println(0)

    // dfs間に合うのか？？
    fun dfs(a: Long, b: Long): Long {
        if (a == x && b == x) return 1
        if (a > x || b > y) return 0L
        return dfs(a + 1, b + 2) + dfs(a + 2, b + 1)
    }

    val answer = dfs(0, 0)

    println(answer)
}