#include<bits/stdc++.h>

using namespace std;


typedef long long ll;

ll M=1e9+7;

int main(){
  ll n,m;
  cin>>n>>m;


  ll ans=(m+1)%M; // d=0

  n--;

  if(n){
    ll t=(m/n)%M;
    
    ans=(ans+t*2)%M; // a0=0 d!=0

    ans=(ans+2*t*((m%n)%M)%M)%M;
    
    n%=M;

    ans=(ans+((t-1)*t%M)*(n)%M)%M;

  }

  cout<<ans<<endl;

  return 0;
}
### Prompt

In CPP, your task is to solve the following problem:

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;


typedef long long ll;

ll M=1e9+7;

int main(){
  ll n,m;
  cin>>n>>m;


  ll ans=(m+1)%M; // d=0

  n--;

  if(n){
    ll t=(m/n)%M;
    
    ans=(ans+t*2)%M; // a0=0 d!=0

    ans=(ans+2*t*((m%n)%M)%M)%M;
    
    n%=M;

    ans=(ans+((t-1)*t%M)*(n)%M)%M;

  }

  cout<<ans<<endl;

  return 0;
}
```