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 <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
#define rep(i, n) for(int i = 0; i < n; i++)
#define REP(i, a, b) for(int i = a; i < b; i++)
#define all(x) x.begin(), x.end()
#define fi first
#define se second
#define pb push_back
#define debug(x) cerr <<__LINE__<< ": " << #x << " -> " << x << endl
#define debug_vec(v) cerr<<__LINE__<<": "<<#v<<" -> ";rep(i,v.size())cerr<<" "<<v[i];cerr<<endl
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
const int INF = (1<<30) - 1;
const ll LINF = (1LL<<60) - 1;
const int mod = 1000000007;

void solve() {
    int n, k;
    cin >> n >> k;
    ll ans = 0;
    REP(i, k, n + 2) {
        ans = (ans + 1LL * (n - i + 1) * i + 1) % mod;
    }
    cout << ans << endl;
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    solve();
    return 0;
}