#include <bits/stdc++.h>
using namespace std;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n)   FOR(i,0,n)
#define pb emplace_back
typedef long long ll;
typedef pair<int,int> pint;

const int mod=1000000007;
int main(){
    ll n,m;
    cin>>n>>m;
    if(n==1){
        cout<<(m+1)%mod<<endl;
        return 0;
    }
    ll k=m/(n-1);
    k%=mod;
    ll mo=(m+1)%mod-((n-1)%mod*k%mod)+mod;
    mo%=mod;
    ll no=((n-1)-mo)%mod+mod;
    no%=mod;
    ll ans=mo*(k+1)%mod*(k+1)%mod+no*k%mod*k%mod;
    cout<<ans%mod<<endl;
    return 0;
}

### Prompt

Create a solution in CPP for 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;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n)   FOR(i,0,n)
#define pb emplace_back
typedef long long ll;
typedef pair<int,int> pint;

const int mod=1000000007;
int main(){
    ll n,m;
    cin>>n>>m;
    if(n==1){
        cout<<(m+1)%mod<<endl;
        return 0;
    }
    ll k=m/(n-1);
    k%=mod;
    ll mo=(m+1)%mod-((n-1)%mod*k%mod)+mod;
    mo%=mod;
    ll no=((n-1)-mo)%mod+mod;
    no%=mod;
    ll ans=mo*(k+1)%mod*(k+1)%mod+no*k%mod*k%mod;
    cout<<ans%mod<<endl;
    return 0;
}

```