Problem:
Score : 400 points

Problem Statement

We have N+1 integers: 10^{100}, 10^{100}+1, ..., 10^{100}+N.

We will choose K or more of these integers. Find the number of possible values of the sum of the chosen numbers, modulo (10^9+7).

Constraints

1 \leq N \leq 2\times 10^5

1 \leq K \leq N+1

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K

Output

Print the number of possible values of the sum, modulo (10^9+7).

Sample Input 1

3 2

Sample Output 1

10

The sum can take 10 values, as follows:

(10^{100})+(10^{100}+1)=2\times 10^{100}+1

(10^{100})+(10^{100}+2)=2\times 10^{100}+2

(10^{100})+(10^{100}+3)=(10^{100}+1)+(10^{100}+2)=2\times 10^{100}+3

(10^{100}+1)+(10^{100}+3)=2\times 10^{100}+4

(10^{100}+2)+(10^{100}+3)=2\times 10^{100}+5

(10^{100})+(10^{100}+1)+(10^{100}+2)=3\times 10^{100}+3

(10^{100})+(10^{100}+1)+(10^{100}+3)=3\times 10^{100}+4

(10^{100})+(10^{100}+2)+(10^{100}+3)=3\times 10^{100}+5

(10^{100}+1)+(10^{100}+2)+(10^{100}+3)=3\times 10^{100}+6

(10^{100})+(10^{100}+1)+(10^{100}+2)+(10^{100}+3)=4\times 10^{100}+6

Sample Input 2

200000 200001

Sample Output 2

1

We must choose all of the integers, so the sum can take just 1 value.

Sample Input 3

141421 35623

Sample Output 3

220280457

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
#include <iostream>
#include <vector>
#include <algorithm>
#include <functional>
#include <cstdlib>
#include <list>
#include <numeric>
#include <map>	
#include <cmath>
#include <string>
#define rep(i,n) for(int i=0;i<n;i++)
#define all(x) (x).begin(),(x).end()
#define sz(x) (int)(x).size()
#define Cout(s) cout<<s<<endl
using namespace std;
typedef long long ll;
string Yes(bool b) { if (b) { return "Yes"; } else { return "No"; } }
string YES(bool b) { if (b) { return "YES"; } else { return "NO"; } }
//================================================
int ketawa(int a) {
	if (a < 10)return a;
	else return a % 10 + ketawa(a / 10);
}

int a = 1, b = 0, c, d, n, k = 0, ans = 0;
string s, t;
int mod = 1000000007;

int main()
{
	ll n, k; cin >> n >> k;
	ll ans = 0;
	for (ll i = k; i < n + 2; i++) {
		ans += (2 * n - i + 1) * i / 2 - i * (i - 1) / 2 + 1;
		ans = ans % mod;
	}
	cout << ans << endl;

	return 0;
}