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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
using namespace std;

#ifdef LOCAL
	#define debug(...) fprintf(stderr, __VA_ARGS__);fflush(stderr);
#else
	#define debug(...) 42
#endif

#define sf scanf
#define pf printf
#define nl printf("\n")
#define ps(s) printf("%s",s.c_str())
#define mem(x,val) memset(x,val,sizeof x)
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend() 
#define sz(x) (int)(x).size()
#define test_case int ___t;scanf("%d", &___t);for(int cs=1;cs<=___t;cs++)

inline string getString(int n){char ch[n+5];sf(" %s",ch);return string(ch);}
inline string getLine(int n)
{char ch[n+5];fgets(ch,n+5,stdin);string s=string(ch);s.pop_back();return s;}

//typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> new_data_set;
typedef long long ll;

const int S=100100;
const int mod=1e9+7;
int main()
{
	ll n,k;
	while(~sf("%lld %lld",&n,&k))
	{
		ll ans=1;
		for(ll i=k;i<=n;i++)
		{
			ll x=n-i;
			x=(x*(x+1))/2;
			ll y=n;
			y=(y*(y+1))/2;
			ll mx=y-x;

			x=i-1;
			ll mn=(x*(x+1))/2;;
			
			ll cur=mx-mn+1;
			ans=(cur+ans)%mod;
		}
		pf("%lld\n",ans);
	}
    return 0;
}