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)

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <cstdio>
#include <vector>
#include <utility>
#include <algorithm>
#include <string>
#include <queue>
#include <stack>
#include <tuple>
#include <list>
#include <set>
#include <map>
#include <functional>
#include <cmath>
#include <cstring>
#include <cstdlib>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> pll;

#define xx first
#define yy second
#define $1 first
#define $2 second

const ll MOD = 1e9 + 7;
ll fast_power(ll base, ll power) {
	ll result = 1;
	while (power > 0) {
		if (power & 1) {
			result = (result*base) % MOD;
		}
		base = (base*base) % MOD;
		power >>= 1;
	}
	return result;
}
void solve() {
	ll n, k; cin >> n >> k;
	ll ans = 0;
	for (ll i = k; i <= n+1; i++) {
		ans += (( (n*(n + 1) / 2) - ((n - i)*(n - i + 1) / 2)) - ((i - 1)*i / 2)) + 1;
		ans %= MOD;
	}
	cout << ans%MOD;
}
int main() {

#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
#endif
	solve();
}