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: Wrong Answer)

#include<bits/stdc++.h> 
using namespace std; 
#define ll long long int 
/* Iterative Function to calculate (x^y)%p 
  in O(log y) */
ll power(ll x, ll y, ll p) 
{ 
    ll res = 1;      // Initialize result 
  
    x = x % p;  // Update x if it is more than or 
                // equal to p 
  
    while (y > 0) 
    { 
        // If y is odd, multiply x with result 
        if (y & 1) 
            res = (res*x) % p; 
  
        // y must be even now 
        y = y>>1; // y = y/2 
        x = (x*x) % p; 
    } 
    return res; 
} 
  
// Returns n^(-1) mod p 
ll modInverse(ll n, ll p) 
{ 
    return power(n, p-2, p); 
} 
  
// Returns nCr % p using Fermat's little 
// theorem. 
ll nCrModPFermat(ll n, ll r, ll p) 
{ 
   // Base case 
   if (r==0) 
      return 1; 
  
    // Fill factorial array so that we 
    // can find all factorial of r, n 
    // and n-r 
    ll fac[n+1]; 
    fac[0] = 1; 
    for (int i=1 ; i<=n; i++) 
        fac[i] = fac[i-1]*i%p; 
  
    return (fac[n]* modInverse(fac[r], p) % p * 
            modInverse(fac[n-r], p) % p) % p; 
} 
  
// Driver program 
int main() 
{ 
    ll n ,r, p = 1000000007; 
  	cin>>n>>r;
    cout<<nCrModPFermat(n+1, r, p); 
    return 0; 
}