#include <bits/stdc++.h>
#define ll long long
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
#define repi(i, a, b) for (ll i = (ll)(a); i < (ll)(b); i++)
#define repm(i, a, b) for (ll i = (ll)(a); i > (ll)(b); i--)
#define all(v) v.begin(), v.end()
#define rall(x) (x).rbegin(), (x).rend()
#define pll pair<long long, long long>
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define vc vector<char>
#define vvc vector<vc>
using namespace std;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<long long>;
using vvll = vector<vll>;
using vb = vector<bool>;
using vvb = vector<vb>;
template <class T>bool chmax(T &a, const T &b){if (a < b){a = b;return 1;}return 0;}
template <class T>bool chmin(T &a, const T &b){if (b < a){a = b;return 1;}return 0;}
//snippets:tmp,cl,cvl,cs,co,coy,con,cov,setp
const ll mod=1e9+7;

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    ll n;
    cin >> n;
    ll m;
    cin >> m;
    ll ans=0;
    if(n!=1){
        ans=2*(m%mod+1)%mod*((m/(n-1))%mod)-((m/(n-1))%mod)*((m/(n-1)+1)%mod)*(n-1)%mod+m+1;
    }else{
        ans=(m+1)%mod;
    }
    cout<<ans%mod<<endl;
}


### Prompt

Construct a cpp code solution to the problem outlined:

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>
#define ll long long
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
#define repi(i, a, b) for (ll i = (ll)(a); i < (ll)(b); i++)
#define repm(i, a, b) for (ll i = (ll)(a); i > (ll)(b); i--)
#define all(v) v.begin(), v.end()
#define rall(x) (x).rbegin(), (x).rend()
#define pll pair<long long, long long>
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define vc vector<char>
#define vvc vector<vc>
using namespace std;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<long long>;
using vvll = vector<vll>;
using vb = vector<bool>;
using vvb = vector<vb>;
template <class T>bool chmax(T &a, const T &b){if (a < b){a = b;return 1;}return 0;}
template <class T>bool chmin(T &a, const T &b){if (b < a){a = b;return 1;}return 0;}
//snippets:tmp,cl,cvl,cs,co,coy,con,cov,setp
const ll mod=1e9+7;

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    ll n;
    cin >> n;
    ll m;
    cin >> m;
    ll ans=0;
    if(n!=1){
        ans=2*(m%mod+1)%mod*((m/(n-1))%mod)-((m/(n-1))%mod)*((m/(n-1)+1)%mod)*(n-1)%mod+m+1;
    }else{
        ans=(m+1)%mod;
    }
    cout<<ans%mod<<endl;
}


```