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
#include <algorithm>
#include <cassert>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <ctime>
#include <deque>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <string>
#include <utility>
#include <vector>

#define REP(i,s,n) for(int i=(int)(s);i<(int)(n);i++)

using namespace std;
typedef long long int ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef pair<int, int> PI;
const ll mod = 1e9 + 7;



int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  ll n, m;
  cin >> n >> m;
  if (n == 1) {
    cout << (m + 1) % mod << "\n";
    return 0;
  }
  ll tot = (m + 1) % mod;
  ll lim = (m + 1) / (n - 1);
  tot *= (2 * lim + 1) % mod;
  tot %= mod;
  ll tmp = (n - 1) % mod;
  tmp *= lim % mod;
  tmp %= mod;
  tmp *= (lim + 1) % mod;
  tmp %= mod;
  tot = (tot + mod - tmp) % mod;
  cout << tot << "\n";
}