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: Wrong Answer)

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<time.h>

#define MOD 1000000007


unsigned long long int modinv(unsigned long long int a){
    long long int b = MOD, u = 1, v = 0;
    while(b > 0){
        long long int t = a/b;
        a -= t*b;
        u -= t*v;
        long long buf = a;
        a = b;
        b = buf;
        buf = u;
        u = v;
        v = buf;
    }
    if(u<0){
        u += MOD;
    }
    u = u%MOD;
    return u;
}

long long fac[510000], finv[510000], inv[510000];

void init(){
    fac[0] = 1;
    fac[1] = 1;
    finv[0] = 1;
    finv[1] = 1;
    inv[1] = 1;
    for(int i=2; i<510000; i++){
        fac[i] = fac[i-1]*i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD/i) % MOD;
        finv[i] = finv[i-1] * inv[i] % MOD;
    }
}

unsigned long long int combination(unsigned long long int n,unsigned long long int r){
    if(r==0||r==n){
        return 1;
    }
    if(r==1){
        return n;
    }
    return (fac[n] * (finv[r] * finv[n-r] % MOD)) % MOD;
}
int main(){
    unsigned long long int x, y;
    scanf("%llu%llu", &x, &y);
    unsigned long long int ans = 0;
    init();
    for(unsigned long long int i=0; i<=x; i++){
        if(2*(x-2*i) == (y-i)){
            ans += combination(i + (x - 2*i), i)%MOD;
            ans = ans%MOD;
        }
    }
    printf("%llu", ans);

    return 0;
}