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 9.2.1) (judge verdict: Runtime Error)

/*
 * FileName:     D_fix
 * CreatedDate:  2020-07-04 01:04:39 +0900
 * LastModified: 2020-07-04 01:17:00 +0900
 */

#include <stdio.h>
#define m 1000001
long int dp[m][m];
int main(void){
    long int x, y;scanf("%ld%ld", &x, &y);
    dp[0][0] = 1;
    for(long int i=0;i<x+1;i++){
        for(long int j=0;j<y+1;j++){
            if(dp[i][j]!=0){
                dp[i+1][j+2] = (dp[i+1][j+2]+dp[i][j])%m;
                dp[i+2][j+1] = (dp[i+2][j+1]+dp[i][j])%m;
            }
        }
    }
    printf("%ld\n",dp[x][y]%m);
    return 0;
}