#include <bits/stdc++.h>
using namespace std;
/*{{{*/  //template
#define rep(i,n) for(int i=0;i<n;i++)
constexpr int INF = numeric_limits<int>::max()/2;
constexpr long long LINF = numeric_limits<long long>::max()/3;
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define all(v) (v).begin(),(v).end()
#define sz(x) (int)(x).size()
#define debug(x) cerr<<#x<<":"<<x<<endl
#define debug2(x,y) cerr<<#x<<","<<#y":"<<x<<","<<y<<endl
//struct fin{ fin(){ cin.tie(0); ios::sync_with_stdio(false); } } fin_;
struct Double{ double d; explicit Double(double x) : d(x){} };
ostream& operator<<(ostream& os,const Double x){ os << fixed << setprecision(20) << x.d; return os; }
template<typename T> ostream& operator<<(ostream& os,const vector<T>& vec){ os << "["; for(const auto& v : vec){ os << v << ","; } os << "]"; return os; }
template<typename T,typename U> ostream& operator<<(ostream& os,const pair<T,U>& p){ os << "(" << p.first << ","<< p.second <<")"; return os; }
template<typename T> ostream& operator<<(ostream& os,const set<T>& st){ os<<"{"; for(T v:st) os<<v<<","; os <<"}"; return os; }
template<typename T,typename U> inline void chmax(T &x,U y){ if(y>x) x = y; }
template<typename T,typename U> inline void chmin(T &x,U y){ if(y<x) x = y; }
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
ll gcd(ll a,ll b){ if(b==0) return a; else return gcd(b,a%b); }
//constexpr double eps = 1e-14; 
constexpr double eps = 1e-10; 
constexpr ll mod = 1e9+7;
const int dx[]={1,0,-1,0} ,dy[] = {0,1,0,-1};
/*}}}*/

ll n,m;

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

    if(n==1){
        cout << (m+1)%mod << endl;
        return 0;
    }

    /*
     * n?????????????´??????? s[n-1] = s[0] + d * (n-1)
     * d??????
     * 0 <= s[0] + d * (n-1) <= m
     * 
     * d = 0 --> m + 1 ??????
     *
     * d > 0 (d < 0 ???????????????????§°???)
     *  --> d?????¶?´? : 0 < d <= m/(n-1) (=ub??¨??????)
     *  --> d??????????????????????????¨????????¨
     *  --> 0 <= s[0] <= m-d(n-1) :: m+1-d(n-1) ??????
     *  --> ?£m+1-d(n-1) (0 < d <= ub)
     *        = (m+1)*ub - (n-1)*(ub*(ub+1))/2 (= a??¨??????)
     *
     *  ????????? m + 1 + 2 * a ????????????????
     */


    ll ub = m/(n-1);
    ll ubmod = ub % mod;

    ll a1 = ((m+1)%mod)*(ub%mod)%mod;    // (m+1)*ub

    ll t1 = (n-1)%mod;                  // (n-1)
    ll t2 = (ubmod*(ubmod+1))/2%mod;    // (ub*(ub+1))/2
    ll a2 = (t1*t2)%mod;
    ll a = (a1-a2+mod) % mod;
    assert(a>=0);
    
    ll b = (m+1)%mod;

    ll ans = (2*a%mod+b)%mod;
    cout << ans << endl;
}
### 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;
/*{{{*/  //template
#define rep(i,n) for(int i=0;i<n;i++)
constexpr int INF = numeric_limits<int>::max()/2;
constexpr long long LINF = numeric_limits<long long>::max()/3;
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define all(v) (v).begin(),(v).end()
#define sz(x) (int)(x).size()
#define debug(x) cerr<<#x<<":"<<x<<endl
#define debug2(x,y) cerr<<#x<<","<<#y":"<<x<<","<<y<<endl
//struct fin{ fin(){ cin.tie(0); ios::sync_with_stdio(false); } } fin_;
struct Double{ double d; explicit Double(double x) : d(x){} };
ostream& operator<<(ostream& os,const Double x){ os << fixed << setprecision(20) << x.d; return os; }
template<typename T> ostream& operator<<(ostream& os,const vector<T>& vec){ os << "["; for(const auto& v : vec){ os << v << ","; } os << "]"; return os; }
template<typename T,typename U> ostream& operator<<(ostream& os,const pair<T,U>& p){ os << "(" << p.first << ","<< p.second <<")"; return os; }
template<typename T> ostream& operator<<(ostream& os,const set<T>& st){ os<<"{"; for(T v:st) os<<v<<","; os <<"}"; return os; }
template<typename T,typename U> inline void chmax(T &x,U y){ if(y>x) x = y; }
template<typename T,typename U> inline void chmin(T &x,U y){ if(y<x) x = y; }
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
ll gcd(ll a,ll b){ if(b==0) return a; else return gcd(b,a%b); }
//constexpr double eps = 1e-14; 
constexpr double eps = 1e-10; 
constexpr ll mod = 1e9+7;
const int dx[]={1,0,-1,0} ,dy[] = {0,1,0,-1};
/*}}}*/

ll n,m;

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

    if(n==1){
        cout << (m+1)%mod << endl;
        return 0;
    }

    /*
     * n?????????????´??????? s[n-1] = s[0] + d * (n-1)
     * d??????
     * 0 <= s[0] + d * (n-1) <= m
     * 
     * d = 0 --> m + 1 ??????
     *
     * d > 0 (d < 0 ???????????????????§°???)
     *  --> d?????¶?´? : 0 < d <= m/(n-1) (=ub??¨??????)
     *  --> d??????????????????????????¨????????¨
     *  --> 0 <= s[0] <= m-d(n-1) :: m+1-d(n-1) ??????
     *  --> ?£m+1-d(n-1) (0 < d <= ub)
     *        = (m+1)*ub - (n-1)*(ub*(ub+1))/2 (= a??¨??????)
     *
     *  ????????? m + 1 + 2 * a ????????????????
     */


    ll ub = m/(n-1);
    ll ubmod = ub % mod;

    ll a1 = ((m+1)%mod)*(ub%mod)%mod;    // (m+1)*ub

    ll t1 = (n-1)%mod;                  // (n-1)
    ll t2 = (ubmod*(ubmod+1))/2%mod;    // (ub*(ub+1))/2
    ll a2 = (t1*t2)%mod;
    ll a = (a1-a2+mod) % mod;
    assert(a>=0);
    
    ll b = (m+1)%mod;

    ll ans = (2*a%mod+b)%mod;
    cout << ans << endl;
}
```