#include <bits/stdc++.h>
// aya_seは起床のプロ
using namespace std;

//@ebi's definition

#define rep(i, a, n) for (int i = (int)(a); i < (int)(n); i++)
#define rrep(i, a, n) for (int i = ((int)(n - 1)); i >= (int)(a); i--)
#define all(v) (v).begin(), (v).end()

using ll = long long;

struct Edge
{
    int to;
    ll weight;
    Edge(int t, ll w) : to(t), weight(w) {}
};

using Graph = vector<vector<Edge>>;
using graph = vector<vector<ll>>;

// xxpoxx's definition
using pll = pair<ll, ll>;
using vl = vector<ll>;
using vll = vector<vl>;
using vpll = vector<pll>;

#define FOR(i, a, b) for (ll i = (a); i < (b); i++)
#define REP(i, n) FOR(i, 0, n)
#define ALL(a) (a).begin(), (a).end()
#define RES(a, n) (a).resize(n)
/***** define constant start *******/
const ll MOD = 1000000007;
const double PI = 2.0 * asin(1.0);
const ll INF = 1L << 60;
string abc = "abcdefghijklmnopqrstuvwxyz";
string ABC = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
const ll MAX = 10000000;
/****** define constant end ********/

// aya_se's definition
vl fac, finv, inv;
void COMinit(ll N)
{
    fac.resize(N);
    finv.resize(N);
    inv.resize(N);
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (ll i = 2; i < N; i++)
    {
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
ll COM(ll n, ll k)
{
    if (n < k)
        return 0;
    if (n < 0 || k < 0)
        return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
ll modinv(ll a, ll m)
{
    ll b = m, u = 1, v = 0;
    while (b)
    {
        ll t = a / b;
        a -= t * b;
        swap(a, b);
        u -= t * v;
        swap(u, v);
    }
    u %= m;
    if (u < 0)
        u += m;
    return u;
}

template <class T>
inline bool chmin(T &a, T b)
{
    if (a > b)
    {
        a = b;
        return true;
    }
    return false;
}

template <class T>
inline bool chmax(T &a, T b)
{
    if (a < b)
    {
        a = b;
        return true;
    }
    return false;
}

int main()
{
    cout << std::fixed << std::setprecision(10);
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    ll N,M;
    cin >>N>>M;
    if(N==1) cout << (M+1)%MOD<<endl;
    else{
        ll ans=(M+1)%MOD;
        if((N-1)>M){
            cout << ans<<endl;
        }else{
            ll X=(M/(N-1))%MOD;
            ll pre1=((M+1)%MOD*X)%MOD;
            ll pre2=((N-1)%MOD);
            ll pre3=((X*(X+1))%MOD*modinv(2,MOD))%MOD;
            ll pre=((pre1-(pre2*pre3)%MOD)%MOD+MOD)%MOD;
            (ans+=pre*2)%MOD;
            cout << ans%MOD << 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>
// aya_seは起床のプロ
using namespace std;

//@ebi's definition

#define rep(i, a, n) for (int i = (int)(a); i < (int)(n); i++)
#define rrep(i, a, n) for (int i = ((int)(n - 1)); i >= (int)(a); i--)
#define all(v) (v).begin(), (v).end()

using ll = long long;

struct Edge
{
    int to;
    ll weight;
    Edge(int t, ll w) : to(t), weight(w) {}
};

using Graph = vector<vector<Edge>>;
using graph = vector<vector<ll>>;

// xxpoxx's definition
using pll = pair<ll, ll>;
using vl = vector<ll>;
using vll = vector<vl>;
using vpll = vector<pll>;

#define FOR(i, a, b) for (ll i = (a); i < (b); i++)
#define REP(i, n) FOR(i, 0, n)
#define ALL(a) (a).begin(), (a).end()
#define RES(a, n) (a).resize(n)
/***** define constant start *******/
const ll MOD = 1000000007;
const double PI = 2.0 * asin(1.0);
const ll INF = 1L << 60;
string abc = "abcdefghijklmnopqrstuvwxyz";
string ABC = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
const ll MAX = 10000000;
/****** define constant end ********/

// aya_se's definition
vl fac, finv, inv;
void COMinit(ll N)
{
    fac.resize(N);
    finv.resize(N);
    inv.resize(N);
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (ll i = 2; i < N; i++)
    {
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
ll COM(ll n, ll k)
{
    if (n < k)
        return 0;
    if (n < 0 || k < 0)
        return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
ll modinv(ll a, ll m)
{
    ll b = m, u = 1, v = 0;
    while (b)
    {
        ll t = a / b;
        a -= t * b;
        swap(a, b);
        u -= t * v;
        swap(u, v);
    }
    u %= m;
    if (u < 0)
        u += m;
    return u;
}

template <class T>
inline bool chmin(T &a, T b)
{
    if (a > b)
    {
        a = b;
        return true;
    }
    return false;
}

template <class T>
inline bool chmax(T &a, T b)
{
    if (a < b)
    {
        a = b;
        return true;
    }
    return false;
}

int main()
{
    cout << std::fixed << std::setprecision(10);
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    ll N,M;
    cin >>N>>M;
    if(N==1) cout << (M+1)%MOD<<endl;
    else{
        ll ans=(M+1)%MOD;
        if((N-1)>M){
            cout << ans<<endl;
        }else{
            ll X=(M/(N-1))%MOD;
            ll pre1=((M+1)%MOD*X)%MOD;
            ll pre2=((N-1)%MOD);
            ll pre3=((X*(X+1))%MOD*modinv(2,MOD))%MOD;
            ll pre=((pre1-(pre2*pre3)%MOD)%MOD+MOD)%MOD;
            (ans+=pre*2)%MOD;
            cout << ans%MOD << endl;
        }
    }
}


```