#include<bits/stdc++.h>
#define ll long long
#define mod 1000000007ll
using namespace std;
ll fact[2000006];
ll bigmod(ll b, ll p)
{
    if(!p)return 1ll;
    ll ret= bigmod(b, p/2);
    ret= (ret*ret)%mod;
    if(p&1ll)return ret= (ret*b)%mod;
    return ret;
}
ll ncr(ll n, ll r)
{
    ll nume= fact[n];
    ll deno= (fact[n-r]*fact[r])%mod;
    return (nume*bigmod(deno, mod-2))%mod;
}
int main()
{
    fact[0]= 1ll;
    for(ll i=1; i<=2000000; i++)
    fact[i]= (fact[i-1]*i)%mod;
    int n, m, cnt= 0;
    scanf("%d %d", &n, &m);
    if(n<m)swap(n, m);
    int tot= (n+m)/3;
    while(n>0 && m>0 && 2*n>m)n--, n--, m--, cnt++;
    if(2*n==m)printf("%lld\n", ncr(tot, cnt));
    else puts("0");
    return 0;
}

### Prompt

Create a solution in Cpp for the following problem:

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.

Examples

Input

3 3


Output

2


Input

2 2


Output

0


Input

999999 999999


Output

151840682

### Response

```cpp
#include<bits/stdc++.h>
#define ll long long
#define mod 1000000007ll
using namespace std;
ll fact[2000006];
ll bigmod(ll b, ll p)
{
    if(!p)return 1ll;
    ll ret= bigmod(b, p/2);
    ret= (ret*ret)%mod;
    if(p&1ll)return ret= (ret*b)%mod;
    return ret;
}
ll ncr(ll n, ll r)
{
    ll nume= fact[n];
    ll deno= (fact[n-r]*fact[r])%mod;
    return (nume*bigmod(deno, mod-2))%mod;
}
int main()
{
    fact[0]= 1ll;
    for(ll i=1; i<=2000000; i++)
    fact[i]= (fact[i-1]*i)%mod;
    int n, m, cnt= 0;
    scanf("%d %d", &n, &m);
    if(n<m)swap(n, m);
    int tot= (n+m)/3;
    while(n>0 && m>0 && 2*n>m)n--, n--, m--, cnt++;
    if(2*n==m)printf("%lld\n", ncr(tot, cnt));
    else puts("0");
    return 0;
}

```