#include <iostream>
#include <algorithm>
#include <numeric>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <stack>
#include <iomanip>
#include <functional>
#include <bitset>
#include <limits>
#include <cstdio>
#include <cmath>
#include <cassert>
#include <random>

#ifdef DEBUG
#include "library/Utility/debug.cpp"
#else
#define debug(...)
#endif

#define rep(i,n) for(int i=0;i<(n);++i)
#define EL '\n'
#define print(i) std::cout << (i) << '\n'
#define all(v) (v).begin(), (v).end()
using lnt = long long;
struct FIO{FIO(){std::cin.tie(0);std::cout.tie(0);std::ios_base::sync_with_stdio(0);std::cout<<std::fixed<<std::setprecision(15);}}fIO;
/*-*/

constexpr lnt MOD = 1e9+7;
struct mint
{
	lnt v;
	mint():v(0){}
	mint(lnt v):v((v+MOD)%MOD){}
	mint operator-()const{ return mint(0) - *this; }
	mint& operator+=(const mint& a){ if((v+=a.v)>=MOD) v-=MOD; return *this; }
	mint& operator-=(const mint& a){ if((v+=MOD-a.v)>=MOD) v-=MOD; return *this; }
	mint& operator*=(const mint& a){ (v*=a.v)%=MOD; return *this; }
	mint& operator/=(const mint& a){ (*this) *= a.inv(); return *this; }
	mint operator+(const mint& a)const{ return mint(*this) += a; }
	mint operator-(const mint& a)const{ return mint(*this) -= a; }
	mint operator*(const mint& a)const{ return mint(*this) *= a; }
	mint operator/(const mint& a)const{ return mint(*this) /= a; }
	bool operator<(const mint& a)const{ return v < a.v; }
	bool operator==(const mint& a)const{ return v == a.v; }
	mint pow(lnt k)const{ mint r(1),t(v); while(k){ if(k&1) r*=t; t*=t; k>>=1; } return r; }
	mint inv()const{ return pow(MOD-2); }
	static mint comb(lnt n, lnt k) { if(n-k<k) k=n-k; mint num(1), dom(1); for(int i=0;i<k;i++) { num*=n-i; dom*=i+1; } return num/dom; }
	static std::vector<mint> construct_comb(int n) {
		std::vector<mint> c(n+1); mint a = 1; c[0] = a; for(int i=1;i<=n;i++) { a = a*mint(n+1-i)/i; c[i] = a; } return c;
	}
	static std::vector<mint> construct_fact(int n) { std::vector<mint> f(n+1,1); for(int i=2;i<=n;i++) f[i]=f[i-1]*i; return f; }
};
std::istream& operator>>(std::istream&i,mint&a){ lnt t; i>>t; a=mint(t); return i; }
std::ostream& operator<<(std::ostream&o,const mint&a){ o<<a.v; return o; }


int main() {
	lnt n,m;
	std::cin >> n >> m;
	if(n==1) {
		print(mint(m+1));
		return 0;
	}
	lnt x = m/(n-1);
	print(mint(m+1)+mint(2)*(mint(x)*mint(m+1))-mint(n-1)*mint(x)*mint(x+1));
}


### 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 <iostream>
#include <algorithm>
#include <numeric>
#include <vector>
#include <string>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <stack>
#include <iomanip>
#include <functional>
#include <bitset>
#include <limits>
#include <cstdio>
#include <cmath>
#include <cassert>
#include <random>

#ifdef DEBUG
#include "library/Utility/debug.cpp"
#else
#define debug(...)
#endif

#define rep(i,n) for(int i=0;i<(n);++i)
#define EL '\n'
#define print(i) std::cout << (i) << '\n'
#define all(v) (v).begin(), (v).end()
using lnt = long long;
struct FIO{FIO(){std::cin.tie(0);std::cout.tie(0);std::ios_base::sync_with_stdio(0);std::cout<<std::fixed<<std::setprecision(15);}}fIO;
/*-*/

constexpr lnt MOD = 1e9+7;
struct mint
{
	lnt v;
	mint():v(0){}
	mint(lnt v):v((v+MOD)%MOD){}
	mint operator-()const{ return mint(0) - *this; }
	mint& operator+=(const mint& a){ if((v+=a.v)>=MOD) v-=MOD; return *this; }
	mint& operator-=(const mint& a){ if((v+=MOD-a.v)>=MOD) v-=MOD; return *this; }
	mint& operator*=(const mint& a){ (v*=a.v)%=MOD; return *this; }
	mint& operator/=(const mint& a){ (*this) *= a.inv(); return *this; }
	mint operator+(const mint& a)const{ return mint(*this) += a; }
	mint operator-(const mint& a)const{ return mint(*this) -= a; }
	mint operator*(const mint& a)const{ return mint(*this) *= a; }
	mint operator/(const mint& a)const{ return mint(*this) /= a; }
	bool operator<(const mint& a)const{ return v < a.v; }
	bool operator==(const mint& a)const{ return v == a.v; }
	mint pow(lnt k)const{ mint r(1),t(v); while(k){ if(k&1) r*=t; t*=t; k>>=1; } return r; }
	mint inv()const{ return pow(MOD-2); }
	static mint comb(lnt n, lnt k) { if(n-k<k) k=n-k; mint num(1), dom(1); for(int i=0;i<k;i++) { num*=n-i; dom*=i+1; } return num/dom; }
	static std::vector<mint> construct_comb(int n) {
		std::vector<mint> c(n+1); mint a = 1; c[0] = a; for(int i=1;i<=n;i++) { a = a*mint(n+1-i)/i; c[i] = a; } return c;
	}
	static std::vector<mint> construct_fact(int n) { std::vector<mint> f(n+1,1); for(int i=2;i<=n;i++) f[i]=f[i-1]*i; return f; }
};
std::istream& operator>>(std::istream&i,mint&a){ lnt t; i>>t; a=mint(t); return i; }
std::ostream& operator<<(std::ostream&o,const mint&a){ o<<a.v; return o; }


int main() {
	lnt n,m;
	std::cin >> n >> m;
	if(n==1) {
		print(mint(m+1));
		return 0;
	}
	lnt x = m/(n-1);
	print(mint(m+1)+mint(2)*(mint(x)*mint(m+1))-mint(n-1)*mint(x)*mint(x+1));
}


```