Problem:
Score : 500 points

Problem Statement

There is a building with n rooms, numbered 1 to n.

We can move from any room to any other room in the building.

Let us call the following event a move: a person in some room i goes to another room j~ (i \neq j).

Initially, there was one person in each room in the building.

After that, we know that there were exactly k moves happened up to now.

We are interested in the number of people in each of the n rooms now. How many combinations of numbers of people in the n rooms are possible?

Find the count modulo (10^9 + 7).

Constraints

All values in input are integers.

3 \leq n \leq 2 \times 10^5

2 \leq k \leq 10^9

Input

Input is given from Standard Input in the following format:

n k

Output

Print the number of possible combinations of numbers of people in the n rooms now, modulo (10^9 + 7).

Sample Input 1

3 2

Sample Output 1

10

Let c_1, c_2, and c_3 be the number of people in Room 1, 2, and 3 now, respectively. There are 10 possible combination of (c_1, c_2, c_3):

(0, 0, 3)

(0, 1, 2)

(0, 2, 1)

(0, 3, 0)

(1, 0, 2)

(1, 1, 1)

(1, 2, 0)

(2, 0, 1)

(2, 1, 0)

(3, 0, 0)

For example, (c_1, c_2, c_3) will be (0, 1, 2) if the person in Room 1 goes to Room 2 and then one of the persons in Room 2 goes to Room 3.

Sample Input 2

200000 1000000000

Sample Output 2

607923868

Print the count modulo (10^9 + 7).

Sample Input 3

15 6

Sample Output 3

22583772

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <stdbool.h>
#define LL long long
#define DB double
#define INF 1145141919810364364
#define PI 3.14159265358979
#define MIN(x,y) ((x)<(y)?(x):(y))
#define MAX(x,y) ((x)<(y)?(y):(x))
#define FOR(i,a,n) for(i=a;i<n;i++)
#define rep(i,n) FOR(i,0,n)
#define MOD 1000000007
//#define MOD 998244353
#define ERR 0.00000001
#define NUM 400010

void swap(LL* a,LL* b){LL tmp=*a;*a=*b;*b=tmp;}
void YN(int a){if(a==1)printf("Yes\n");if(a==0)printf("No\n");}
LL lpow(LL a,LL n){LL ans=pow(a,n)*10;if(ans%10<5)return ans/10;return ans/10+1;}
LL dup(LL a,LL d){if(a%d==0)return a/d;else return a/d+1;}
LL GCD(LL a,LL b){LL c=1,tmp=MAX(a,b);b=MIN(a,b);a=tmp;while(c!=0){c=a%b;a=b;b=c;}return a;}
LL LCM(LL a,LL b){return a*b/GCD(a,b);}
int lcmp(const void* a,const void* b){if(*(LL*)a>*(LL*)b)return 1;if(*(LL*)a<*(LL*)b)return -1;return 0;}
int lcmpr(const void* a,const void* b){if(*(LL*)a>*(LL*)b)return -1;if(*(LL*)a<*(LL*)b)return 1;return 0;}
int ccmp(const void* a,const void* b){return *(char*)a-*(char*)b;}
int ccmpr(const void* a,const void* b){return *(char*)b-*(char*)a;}
int scmp(const void* a,const void* b){return strcmp((char*)a,(char*)b);}
int scmpr(const void* a,const void* b){return strcmp((char*)b,(char*)a);}
LL mod(LL a,LL m){if(a<0)return a%m+m;else return a%m;}
LL DIV(LL a,LL d){LL m=MOD,x=1,y=0,k;while(m){k=d/m;d-=k*m;swap(&m,&d);x-=k*y;swap(&x,&y);}return mod(a*mod(x,MOD),MOD);}
LL FAC(LL a){LL i,ans=1;FOR(i,1,a+1){ans*=i;if(MOD>0&&ans>MOD)ans%=MOD;}return ans;}
LL POW(LL a,LL n){LL ans=1;while(n>0){if(n&1)ans=ans*a%MOD;a=a*a%MOD;n>>=1;}return ans;}
LL fact[NUM],finv[NUM],inv[NUM];
void comi(){LL i;fact[0]=fact[1]=1;finv[0]=finv[1]=1;inv[1]=1;FOR(i,2,NUM){fact[i]=fact[i-1]*i%MOD;inv[i]=MOD-inv[MOD%i]*(MOD/i)%MOD;finv[i]=finv[i-1]*inv[i]%MOD;}}
LL com(LL n,LL k){if(n<k||n<0||k<0)return 0;return fact[n]*(finv[k]*finv[n-k]%MOD)%MOD;}

int main(){
  comi();
  LL n,k,i,ans=0;
  scanf("%lld%lld",&n,&k);
  ans=com(n*2-1,n-1);
  FOR(i,k+1,n)ans=mod(ans-com(n,i)*com(n-1,i),MOD);

  printf("%lld\n",ans);
	return 0;
}