#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

int main() {
  ll n,m,mod=1000000007;
  cin >> n >> m;
  if(n==1) {
    cout << (m+1)%mod << endl;
    return 0;
  }
  m++;
  ll k=ceil((double)m/(n-1));
  n%=mod;
  m%=mod;
  ll x=(2+(k-1)*2)/2%mod*(k%mod)%mod;
  ll y=((2+(k-1)*2)/2%mod*(k%mod)%mod-(2+(k-2)*2)/2%mod*((k-1+mod)%mod)%mod+mod)%mod;
  ll ans=(x*((n-1+mod)%mod)%mod-y*(k%mod*((n-1+mod)%mod)%mod-m+mod)%mod+mod)%mod;
  cout << ans << endl;
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
typedef long long ll;

int main() {
  ll n,m,mod=1000000007;
  cin >> n >> m;
  if(n==1) {
    cout << (m+1)%mod << endl;
    return 0;
  }
  m++;
  ll k=ceil((double)m/(n-1));
  n%=mod;
  m%=mod;
  ll x=(2+(k-1)*2)/2%mod*(k%mod)%mod;
  ll y=((2+(k-1)*2)/2%mod*(k%mod)%mod-(2+(k-2)*2)/2%mod*((k-1+mod)%mod)%mod+mod)%mod;
  ll ans=(x*((n-1+mod)%mod)%mod-y*(k%mod*((n-1+mod)%mod)%mod-m+mod)%mod+mod)%mod;
  cout << ans << endl;
  return 0;
}

```