#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<b;i++)
#define rrep(i,a,b) for(int i=a;i>=b;i--)
#define fore(i,a) for(auto &i:a)
#pragma GCC optimize ("-O3")
using namespace std; void _main(); int main() { cin.tie(0); ios::sync_with_stdio(false); _main(); }
//---------------------------------------------------------------------------------------------------
template<int MOD> struct ModInt {
    static const int Mod = MOD; unsigned x; ModInt() : x(0) { }
    ModInt(signed sig) { x = sig < 0 ? sig % MOD + MOD : sig % MOD; }
    ModInt(signed long long sig) { x = sig < 0 ? sig % MOD + MOD : sig % MOD; }
    int get() const { return (int)x; }
    ModInt &operator+=(ModInt that) { if ((x += that.x) >= MOD) x -= MOD; return *this; }
    ModInt &operator-=(ModInt that) { if ((x += MOD - that.x) >= MOD) x -= MOD; return *this; }
    ModInt &operator*=(ModInt that) { x = (unsigned long long)x * that.x % MOD; return *this; }
    ModInt &operator/=(ModInt that) { return *this *= that.inverse(); }
    ModInt operator+(ModInt that) const { return ModInt(*this) += that; }
    ModInt operator-(ModInt that) const { return ModInt(*this) -= that; }
    ModInt operator*(ModInt that) const { return ModInt(*this) *= that; }
    ModInt operator/(ModInt that) const { return ModInt(*this) /= that; }
    ModInt inverse() const { long long a = x, b = MOD, u = 1, v = 0;
        while (b) { long long t = a / b; a -= t * b; std::swap(a, b); u -= t * v; std::swap(u, v); }
        return ModInt(u); }
    bool operator==(ModInt that) const { return x == that.x; }
    bool operator!=(ModInt that) const { return x != that.x; }
    ModInt operator-() const { ModInt t; t.x = x == 0 ? 0 : Mod - x; return t; }
};
template<int MOD> ostream& operator<<(ostream& st, const ModInt<MOD> a) { st << a.get(); return st; };
template<int MOD> ModInt<MOD> operator^(ModInt<MOD> a, unsigned long long k) {
    ModInt<MOD> r = 1; while (k) { if (k & 1) r *= a; a *= a; k >>= 1; } return r; }
typedef ModInt<1000000007> mint;

typedef long long ll;
typedef long double ld;
#define INF 1LL<<60
ld lginf = -1;
ll mul(ll a, ll b) {
    if (lginf < 0) lginf = log10l(INF); ld aa = log10l(a), bb = log10l(b);
    if (lginf <= aa + bb) return INF; return a * b;
}
/*---------------------------------------------------------------------------------------------------
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ll N, M;
int mod = 1000000007;
//--------------------------------------------------------------------------------------------------
void _main() {
    cin >> N >> M;

    if (N == 1) {
        int ans = (M + 1) % mod;
        cout << ans << endl;
        return;
    }

    ll ok = 0, ng = M + 1;
    while (ok + 1 != ng) {
        ll d = (ok + ng) / 2;
        ll b = mul(N - 1, d);
        if (M < b) ng = d;
        else ok = d;
    }

    // (M + 1) * ok - sigma(0 + (N - 1) * 1 + ... + (N - 1) * ok)

    mint ans = mint((M + 1) % mod) * mint(ok % mod);
    ans -= mint(ok%mod) / 2 * (mint((N - 1) % mod) + mint((N - 1) % mod) * mint(ok % mod));
    ans *= 2;
    ans += (M + 1) % mod;
    cout << ans << endl;
}
### 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>
#define rep(i,a,b) for(int i=a;i<b;i++)
#define rrep(i,a,b) for(int i=a;i>=b;i--)
#define fore(i,a) for(auto &i:a)
#pragma GCC optimize ("-O3")
using namespace std; void _main(); int main() { cin.tie(0); ios::sync_with_stdio(false); _main(); }
//---------------------------------------------------------------------------------------------------
template<int MOD> struct ModInt {
    static const int Mod = MOD; unsigned x; ModInt() : x(0) { }
    ModInt(signed sig) { x = sig < 0 ? sig % MOD + MOD : sig % MOD; }
    ModInt(signed long long sig) { x = sig < 0 ? sig % MOD + MOD : sig % MOD; }
    int get() const { return (int)x; }
    ModInt &operator+=(ModInt that) { if ((x += that.x) >= MOD) x -= MOD; return *this; }
    ModInt &operator-=(ModInt that) { if ((x += MOD - that.x) >= MOD) x -= MOD; return *this; }
    ModInt &operator*=(ModInt that) { x = (unsigned long long)x * that.x % MOD; return *this; }
    ModInt &operator/=(ModInt that) { return *this *= that.inverse(); }
    ModInt operator+(ModInt that) const { return ModInt(*this) += that; }
    ModInt operator-(ModInt that) const { return ModInt(*this) -= that; }
    ModInt operator*(ModInt that) const { return ModInt(*this) *= that; }
    ModInt operator/(ModInt that) const { return ModInt(*this) /= that; }
    ModInt inverse() const { long long a = x, b = MOD, u = 1, v = 0;
        while (b) { long long t = a / b; a -= t * b; std::swap(a, b); u -= t * v; std::swap(u, v); }
        return ModInt(u); }
    bool operator==(ModInt that) const { return x == that.x; }
    bool operator!=(ModInt that) const { return x != that.x; }
    ModInt operator-() const { ModInt t; t.x = x == 0 ? 0 : Mod - x; return t; }
};
template<int MOD> ostream& operator<<(ostream& st, const ModInt<MOD> a) { st << a.get(); return st; };
template<int MOD> ModInt<MOD> operator^(ModInt<MOD> a, unsigned long long k) {
    ModInt<MOD> r = 1; while (k) { if (k & 1) r *= a; a *= a; k >>= 1; } return r; }
typedef ModInt<1000000007> mint;

typedef long long ll;
typedef long double ld;
#define INF 1LL<<60
ld lginf = -1;
ll mul(ll a, ll b) {
    if (lginf < 0) lginf = log10l(INF); ld aa = log10l(a), bb = log10l(b);
    if (lginf <= aa + bb) return INF; return a * b;
}
/*---------------------------------------------------------------------------------------------------
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??????????????? ??§?????§ ???????´<_??? ?????? Welcome to My Coding Space!
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?????????????????????????????? ???  |???|     
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---------------------------------------------------------------------------------------------------*/


ll N, M;
int mod = 1000000007;
//--------------------------------------------------------------------------------------------------
void _main() {
    cin >> N >> M;

    if (N == 1) {
        int ans = (M + 1) % mod;
        cout << ans << endl;
        return;
    }

    ll ok = 0, ng = M + 1;
    while (ok + 1 != ng) {
        ll d = (ok + ng) / 2;
        ll b = mul(N - 1, d);
        if (M < b) ng = d;
        else ok = d;
    }

    // (M + 1) * ok - sigma(0 + (N - 1) * 1 + ... + (N - 1) * ok)

    mint ans = mint((M + 1) % mod) * mint(ok % mod);
    ans -= mint(ok%mod) / 2 * (mint((N - 1) % mod) + mint((N - 1) % mod) * mint(ok % mod));
    ans *= 2;
    ans += (M + 1) % mod;
    cout << ans << endl;
}
```