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++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include <iostream>
#define rep(i,j,k) for(int i=j;(i)<(k);(i)++)
using namespace std;
#define mod 1000000007
#define mmax 100000
typedef long long ll;
ll f[mmax], rf[mmax];
ll inv(ll x) {
    ll res = 1;
    ll k = mod - 2;
    ll y = x;
    while (k) {
        if (k & 1) res = (res * y) % mod;
        y = (y * y) % mod;
        k /= 2;
    }
    return res;
}
void init() {
    f[0] = 1;
    rep(i, 1, mmax) f[i] = (f[i - 1] * i) % mod;
    rep(i, 0, mmax) rf[i] = inv(f[i]);
}
//------------------------------------------------------------------
ll C(int n, int k) {
    init();
    ll a = f[n]; // = n!
    ll b = rf[n-k]; // = (n-k)!
    ll c = rf[k]; // = k!
    
    ll bc = (b * c) % mod;
    
    return (a * bc) % mod;
}
int main(void){
    int x,y;
    cin >> x >> y;
    if((x+y)%3 != 0){
        printf("0\n");
    }else{
        int layer = (x+y)/3;
        int left =(x+y)-layer;
        left -= x;
        //printf("%d %d\n",layer,left);
        printf("%lld\n",C(layer,left));
        
    }
}