#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<stack>
using namespace std;

#define int long long
#define REP(i,n) for(int i=0;(i)<(n);(i)++)
#define ll long long

static const int MOD = 1000000007;

int n, m;
int ns, f, d;
int ans;

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

	if (n == 1) {
		ns = 0;
	}
	else ns = ((m + 1) / (n - 1)) % MOD;
	f = (m + 1 - (n - 1)) % MOD;
	d = -(n - 1) % MOD;

	ans = (ns*((2 * f + ((ns - 1)*d) % MOD) % MOD) % MOD + m + 1) % MOD;

	cout << ans << endl;
	return 0;
}
### Prompt

Your task is to create a Cpp solution to 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
#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<stack>
using namespace std;

#define int long long
#define REP(i,n) for(int i=0;(i)<(n);(i)++)
#define ll long long

static const int MOD = 1000000007;

int n, m;
int ns, f, d;
int ans;

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

	if (n == 1) {
		ns = 0;
	}
	else ns = ((m + 1) / (n - 1)) % MOD;
	f = (m + 1 - (n - 1)) % MOD;
	d = -(n - 1) % MOD;

	ans = (ns*((2 * f + ((ns - 1)*d) % MOD) % MOD) % MOD + m + 1) % MOD;

	cout << ans << endl;
	return 0;
}
```