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 C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
//https://atcoder.jp/contests/abc145/tasks/abc145_d
const long int modo = 1000000007;
const long int modo2 = 1000000005;

long long int powMod(long long int a, long long int b){
    long long int x=1, y = a;
    while(b > 0){
        if(b%2==1){
            x = (x*y);
            if(x > modo) 
                x %= modo;
        }
        y = (y * y);
        if(y > modo) 
            y %= modo;
        b/=2;
    }
    return x;
}
 
 
long long int inverseEuler(long long int n){
    return powMod(n, modo2);
}

 
long long int C(long long int n, long long int r){
    long long int i, *f = malloc((n + 1) * sizeof(long long int));
    f[0] = 1;
    f[1] = 1;
    for(i = 2;i<=n;i++){
        f[i] = (f[i-1]*i);
        if(f[i] > modo)
            f[i] %= modo;
    }
 
    return (f[n]*((inverseEuler(f[r]) * inverseEuler(f[n-r]))%modo))%modo;
}
int main() {
	long int x, y, n1, N;
	scanf("%ld %ld\n", &x, &y);
	n1 =  (x < y) ? (2 * x - y) / 3 : (2 * y - x) / 3;
	N = y + x;
	if((N % 3 != 0) || n1 < 0 || n1 > N)
	    printf("0\n");
	else{
	    N /= 3;
	    printf("%lld\n", C(N, n1));
	}
	return 0;
}