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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

#define all(a) (a).begin(), (a).end()
#define rall(a) (a).rbegin(), (a).rend()
#define MIN(a) *min_element(all(a))
#define MAX(a) *max_element(all(a))
#define SUM(a) accumulate(all(a), 0LL)
#define REP(i, n) for(int (i)=0; (i)<(n); (i)++)
#define RREP(i, n) for(int (i)=(n)-1; (i)>=0; (i)--)
#define FOR(i, m, n) for(int (i)=(m); (i)<(n); i++)
#define FORR(i, m, n) for(int (i)=(n)-1; (i)>=(m); i--)
#define debug(x) cerr << #x << " = " << x << endl
//#define int long long

typedef long long ll;
//ll const MOD = 998244353;
ll const MOD = 1e9+7;
int const inf = 1e9;
ll const INF = 1e18;

inline void ios_(){cin.tie(0); ios::sync_with_stdio(false);}
template<typename T> int size(const T& a){return (int)a.size();}
template<typename T> T Div(T a, T b){return (a + b - 1) / b;}
template<typename T> bool chmin(T& a, const T& b){if(a > b){a = b; return true;} return false;}
template<typename T> bool chmax(T& a, const T& b){if(a < b){a = b; return true;} return false;}

class Combination{
public:
    int N;
    ll mod;
    vector<ll> fact;

    Combination(int n, ll m) : N(n), mod(m), fact(n+1){
        fact[0] = 1;
        for(ll i=1; i<=N; i++){
            fact[i] = i * fact[i-1] % mod;
        }
    }

    ll POW(ll a, ll b){
        ll res = 1;
        while(b > 0){
            if(b & 1){
                res = res * a % mod;
            }
            a = a * a % mod;
            b >>= 1;
        }
        return res;
    }

    ll combin(int n, int k){
        if(n < 0 || k < 0 || n < k){
            return 1;
        }
        return fact[n] * POW(fact[k] * fact[n-k] % mod, mod - 2) % mod;
    }

    ll moddiv(ll a, ll b){
        return a % MOD * POW(b, mod - 2) % MOD;
    }
};

signed main(){
    ll n, k;
    cin >> n >> k;

    Combination C(2*n, MOD);
    ll ans = 0;

    REP(i, n){
        if(i > k) break;
        ll cnt = C.combin(n, i);
        cnt *= C.combin(n-1, i);
        cnt %= MOD;
        ans += cnt;
        ans %= MOD;
    }

    cout << ans << endl;

    return 0;
}