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>
#include <math.h>
#include <numeric>
#include <vector>
#include <map>
#include <algorithm>
#include <queue>
#include <tuple>
#include <functional>

#define PI 3.14159265359
#define INF 1e9
#define LINF 1e18
#define rep(i,n) for(long long  i = 0; i < n; i++)
#define rep2(i,a,n) for(long long  i = a; i < n; i++)
#define in(a) cin >> a
#define out(a) cout << a << endl

using namespace std;

typedef long long ll;
typedef pair<ll, ll> PLL;

int main() 
{
    ll x,y;
    cin >> x >> y;
    vector<vector<ll> > g(x+1, vector<ll>(y+1, 0));
    
    g[0][0] = 1;
    g[x][y] = 0;

    rep(i,x+1) {
        rep(j,y+1) {
                
            if((i-1 > -1) && (j-2 > -1)) { 
                //cout << "i," << i << ",j," << j << endl;
                if(g[i-1][j-2] != 0) {
                    g[i][j] += g[i-1][j-2];
                    //cout << g[i][j] << ","; 
                    //cout << g[i-1][j-2] << endl; 
                }
            }
            
            if((i-2 > -1) && (j-1 > -1)) { 
                //cout << "i," << i << ",j," << j << endl;
                if(g[i-2][j-1] != 0) {
                    g[i][j] += g[i-2][j-1];
                    //cout << g[i][j] << ","; 
                    //cout << g[i-2][j-1] << endl; 
                }
            }
        }
    }
    
    cout << g[x][y] % (1000000000+7)<< endl;

    return 0;
}