#include<bits/stdc++.h>
using namespace std;
int main(){
  unsigned long long n,m,ans,mod=1e9+7;
  cin>>n>>m;
  if(n==1)cout<<(m+1)%mod<<endl;
  else if(m==0)cout<<1<<endl;
  else{
    ans=( ((m/(n-1))%mod) * ((m-n+3+m%(n-1))%mod) )%mod+1+m;
    // ans=((m-(n-1)+2+m%(n-1))%mod*(m/(n-1))%mod)+m+1;
  cout<<ans%mod<<endl;
  }
  return 0;
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

B: Ebi-chan and Integer Sequences-

problem

Ebi-chan likes sequences. I especially like arithmetic progressions. This time, I decided to create a sequence that meets the following conditions.

* Arithmetic progression of length n
* When the i-th element of the sequence is defined as s_i, all s_i (1 \ leq i \ leq n) are integers that satisfy 0 \ leq s_i \ leq m.



How many sequences are there that satisfy the above conditions? However, the answer can be very large, so print the remainder after dividing by 10 ^ 9 + 7.

Input format

Input is given in one line.


n m

n represents the length of the sequence.

Constraint

* 1 \ leq n \ leq 10 ^ {15}
* 0 \ leq m \ leq 10 ^ {15}



Output format

Divide the number of arithmetic progressions that satisfy the condition by 10 ^ 9 + 7 and output the remainder in one row.

Input example 1


3 9

Output example 1


50

Input example 2


10000000000 10000000000

Output example 2


999999942

Note that the input does not fit in a 32-bit integer.





Example

Input

3 9


Output

50

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
int main(){
  unsigned long long n,m,ans,mod=1e9+7;
  cin>>n>>m;
  if(n==1)cout<<(m+1)%mod<<endl;
  else if(m==0)cout<<1<<endl;
  else{
    ans=( ((m/(n-1))%mod) * ((m-n+3+m%(n-1))%mod) )%mod+1+m;
    // ans=((m-(n-1)+2+m%(n-1))%mod*(m/(n-1))%mod)+m+1;
  cout<<ans%mod<<endl;
  }
  return 0;
}
```