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;
#define ll long long
#define n_l '\n'

#ifndef ONLINE_JUDGE
#define dbg(...) cout << "[" << #__VA_ARGS__ << "]: "; cout << to_string(__VA_ARGS__) << endl
template <typename T, size_t N> int SIZE(const T (&t)[N]){ return N; } template<typename T> int SIZE(const T &t){ return t.size(); } string to_string(const string s, int x1=0, int x2=1e9){ return '"' + ((x1 < s.size()) ? s.substr(x1, x2-x1+1) : "") + '"'; } string to_string(const char* s) { return to_string((string) s); } string to_string(const bool b) { return (b ? "true" : "false"); } string to_string(const char c){ return string({c}); } template<size_t N> string to_string(const bitset<N> &b, int x1=0, int x2=1e9){ string t = ""; for(int __iii__ = min(x1,SIZE(b)),  __jjj__ = min(x2, SIZE(b)-1); __iii__ <= __jjj__; ++__iii__){ t += b[__iii__] + '0'; } return '"' + t + '"'; } template <typename A, typename... C> string to_string(const A (&v), int x1=0, int x2=1e9, C... coords); int l_v_l_v_l = 0, t_a_b_s = 0; template <typename A, typename B> string to_string(const pair<A, B> &p) { l_v_l_v_l++; string res = "(" + to_string(p.first) + ", " + to_string(p.second) + ")"; l_v_l_v_l--; return res; } template <typename A, typename... C> string to_string(const A (&v), int x1, int x2, C... coords) { int rnk = rank<A>::value; string tab(t_a_b_s, ' '); string res = ""; bool first = true; if(l_v_l_v_l == 0) res += n_l; res += tab + "["; x1 = min(x1, SIZE(v)), x2 = min(x2, SIZE(v)); auto l = begin(v); advance(l, x1); auto r = l; advance(r, (x2-x1) + (x2 < SIZE(v))); for (auto e = l; e != r; e = next(e)) { if (!first) { res += ", "; } first = false; l_v_l_v_l++; if(e != l){ if(rnk > 1) { res += n_l; t_a_b_s = l_v_l_v_l; }; } else{ t_a_b_s = 0; } res += to_string(*e, coords...); l_v_l_v_l--; } res += "]"; if(l_v_l_v_l == 0) res += n_l; return res; } void dbgm(){;} template<typename Heads, typename... Tails> void dbgm(Heads H, Tails... T){ cout << to_string(H) << " | "; dbgm(T...); } 
#define dbgm(...) cout << "[" << #__VA_ARGS__ << "]: "; dbgm(__VA_ARGS__); cout << endl
#endif

#define pb push_back
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define sz(x) ((int)(x).size())
#define forn(i,n) for(int i=0;i<n;i++)
#define for1(i,n) for(int i=1;i<=n;i++)
typedef vector<int> vi;
typedef vector<ll> vl;
typedef pair<int,int> pii;
typedef double db;
typedef vector<vi> vvi;
const ll mod=1000000007;
const ll N=200005; //check this
int dx[] = {-1, 0, 1, 0, 1, 1, -1, -1};
int dy[] = {0, -1, 0, 1, 1, -1, 1, -1};

ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll powb(ll a,ll b) {ll res=1;a ; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a;a=a*a;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}

int main(){
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

    #ifndef ONLINE_JUDGE
    freopen("inputf.in", "r", stdin);
    freopen("outputf.in", "w", stdout);
    #endif

    int t=1;
    //cin>>t;
    while(t--){
    	ll n,k;cin>>n>>k;
    	ll ans=0;
    	for(int i=k;k<=n+1;k++){
    		ll mn=k*(k-1)/2;
    		ll mx=n*(n+1)/2-(n-k)*(n-k+1)/2;
    		// cout<<mn<<" "<<mx<<"\n";
    		(ans+=max(0LL,mx-mn+1))%=mod;
    	}
    	cout<<ans;

    }
}