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: Accepted)

#include<iostream>
#include<vector>
#include<string>
#include<cmath>
#include<algorithm>
#include<numeric>
#include<map>
#include<unordered_map>
#include<queue>
#include<deque>
 
using namespace std;
using ll=long long;
#define rep(i,n)  for(ll i=0;i<n;++i)
#define all_map(itr,mp) for(auto itr=mp.begin();itr!=mp.end();itr++)
#define ALL(a) (a).begin(),(a).end()
 
//comb　nCk
#define MAX 1000009
#define MOD 1000000007
ll fac[MAX], finv[MAX], inv[MAX];
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; 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;
    }
}
ll COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
ll x, y;
 
int main(){
    cin >> x >> y;
    COMinit();
    if(x&1){
        if((y-(x+3)/2)%3!=0 || x*2<y || y<(x+3)/2)cout << 0;
        else cout << COM((x+1)/2+(y-(x+3)/2)/3, (y-(x+3)/2)*2/3+1);
    }
    else {
        if((y-x/2)%3!=0 || x*2<y || y<x/2)cout << 0;
        else cout << COM(x/2+(y-x/2)/3, (y-x/2)*2/3);
    }
}