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;
#ifndef LOCAL
#define endl '\n'
#endif

#define fr(i, a, b) for(int i = a; i <= b; i++)
#define rf(i, a, b) for(int i = a; i >= b; i--)
#define pf push_front
#define pb push_back
#define fi first
#define se second
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define sz(x) (int)x.size()
#define rsz resize()
#define lb lower_bound
#define ub upper_bound
#define br cout << endl

typedef long long ll;
typedef long double f80;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

int pct(int x) { return __builtin_popcount(x); }
int pct(ll x) { return __builtin_popcountll(x); } 
int bit(int x) { return 31 - __builtin_clz(x); } // floor(log2(x))
int bit(ll x) { return 63 - __builtin_clzll(x); } // floor(log2(x))
int cdiv(int a, int b) { return a / b + !(a < 0 || a % b == 0); }
ll cdiv(ll a, ll b) { return a / b + !(a < 0 || a % b == 0); }

template<typename T>
void leftShift(vector<T> &v, ll k) { k %= sz(v); if(k < 0) k += sz(v); rotate(v.begin(), v.begin() + k, v.end()); }
template<typename T>
void rightShift(vector<T> &v, ll k) { leftShift(v, sz(v) - k); }

mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());

ll rand(ll l, ll r){
	uniform_int_distribution<ll> uid(l, r);
	return uid(rng);
}

inline int nxt() {
	int x;
	cin >> x;
	return x;
}

inline int nxtll() {
	ll x;
	cin >> x;
	return x;
}

void pr() {}
void sc() {}

template <typename Head, typename... Tail>
void pr(Head H, Tail... T) { cout << H << " "; pr(T...); }

template <typename Head, typename... Tail>
void sc(Head &H, Tail &... T) { cin >> H; sc(T...); }

#ifdef LOCAL
#define debug(...) cerr << "[L:" << __LINE__ << "][" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 42
#endif

#ifndef LOCAL
string to_string(__int128 x) {
	string s = "";
	bool neg = 0;
	if(x < 0) { s += "-"; neg = 1; }
	if(!x) s += '0';
	while(x) {
		int rem = x % 10;
		s += to_string(rem);
		x /= 10;
	}
	reverse(s.begin() + neg, s.end());
	return s;
}
#endif

const int mod = 1e9 + 7; // 998244353;

int pwr(int a,int b) {
	int ans = 1;
	while(b) {
		if(b & 1) ans = (ans * 1LL * a) % mod;
		a = (a * 1LL * a) % mod;
		b >>= 1;
	}
	return ans;
}

/*
	Lookout for overflows!!
	Check array sizes!!
	Clear before test cases!!
	Use the correct modulo!!
	Check for corner cases!!
	Are you forgetting something?!
*/

const int N = 2e5 + 5;
int fact[N], ifact[N];

int ncr(int n,int k) {
	int ans = (fact[n] * 1LL * ifact[n - k]) % mod;
	return (ans * 1LL * ifact[k]) % mod;
}

ll sum(ll l, ll r) {
	l--;
	return r * (r + 1) / 2 - l * (l + 1) / 2;
}

void solve() {
	int n = nxt(), k = nxt();
	fact[0] = 1;
	ifact[0] = 1;
	fr(i, 1, N - 1) {
		fact[i] = (i * 1LL * fact[i - 1]) % mod;
		ifact[i] = pwr(fact[i], mod - 2);
	}
	int ans = 0;
	fr(i, k, n + 1) {
		ll maxsum = sum(n - i + 1, n);
		ll minsum = sum(0, i - 1);
		ans = (ans + maxsum - minsum + 1) % mod;
	}
	cout << ans;
}

signed main() {
	ios :: sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int t;
	// cin >> t;
	t = 1;
	for(int tt = 1; tt <= t; tt++) {
		solve();
	}
	return 0;
}