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: Wrong Answer)

#include <bits/stdc++.h>


#define f(i,n) for(int i=0;i<(n);i++)


#define inf (int)(3e18)
#define int long long
#define mod (int)(1000000007)

#define intt long long
#define P pair<int,int>

using namespace std;

//Library
//モッドパウ
int modpow(int x, int y, int m = mod) {
    int res = 1;
    while (y) {
        if (y % 2) {
            res *= x;
            res %= m;
        }
        x = x * x % mod;
        y /= 2;
    }
    return res;
}

int mypow(int x, int y) {
    int res = 1;
    while (y) {
        if (y % 2) {
            res *= x;
        }
        x = x * x;
        y /= 2;
    }
    return res;
}
//is the number (x) a prime number?
bool prime(int x) {
    for (int i = 2; i * i <= x; i++) {
        if (!(x % i)) {
            return false;
        }
    }
    return true;
}

double kyori(pair<int, int> f, pair<int, int> s) {
    double ans = 0;
    double t = fabs(f.first - s.first);
    double y = fabs(f.second - s.second);
    ans = sqrt(t * t + y * y);
    return ans;
}

//saidai-kouyakusuu
inline int gcd(int x, int y) {
    if (!y) {
        return x;
    }
    return gcd(y, x % y);
}

//Union-Find Tree

class Union_Find {
    vector<int> par;
    vector<int> rankmy;
public:
    Union_Find(int size) {
        par = vector<int>(size);
        rankmy = vector<int>(size);
        for (int i = 0; i < size; i++) {
            par[i] = i;
        }
    }

    int find(int x) {
        if (par[x] == x) {
            return x;
        }
        return par[x] = find(par[x]);
    }

    void unite(int x, int y) {
        x = find(x);
        y = find(y);
        if (x == y) {
            return;
        }
        if (rankmy[x] < rankmy[y]) {
            par[x] = y;
        }
        else {
            par[y] = x;
            if (rankmy[x] == rankmy[y]) {
                rankmy[x]++;
            }
        }
    }

    bool same(int x, int y) {
        return find(x) == find(y);
    }

};

//Union-Find-End

//SegTree
template <class T>
class SegTree {
    int n;                       // 葉の数
    vector<T> data;              // データを格納するvector
    T def;                       // 初期値かつ単位元
    function<T(T, T)> operation; // 区間クエリで使う処理
    function<T(T, T)> update;    // 点更新で使う処理

    // 区間[a,b)の総和。ノードk=[l,r)に着目している。
    T _query(int a, int b, int k, int l, int r) {
        if (r <= a || b <= l) return def; // 交差しない
        if (a <= l && r <= b)
            return data[k]; // a,l,r,bの順で完全に含まれる
        else {
            T c1 = _query(a, b, 2 * k + 1, l, (l + r) / 2); // 左の子
            T c2 = _query(a, b, 2 * k + 2, (l + r) / 2, r); // 右の子
            return operation(c1, c2);
        }
    }

public:
    // _n:必要サイズ, _def:初期値かつ単位元, _operation:クエリ関数,
    // _update:更新関数
    SegTree(size_t _n, T _def, function<T(T, T)> _operation,
function<T(T, T)> _update)
        : def(_def), operation(_operation), update(_update) {
        n = 1;
        while (n < _n) {
            n *= 2;
        }
        data = vector<T>(2 * n - 1, def);
    }

    // 場所i(0-indexed)の値をxで更新
    void change(int i, T x) {
        i += n - 1;
        data[i] = update(data[i], x);
        while (i > 0) {
            i = (i - 1) / 2;
            data[i] = operation(data[i * 2 + 1], data[i * 2 + 2]);
        }
    }

    // [a, b)の区間クエリを実行
    T query(int a, int b) {
        return _query(a, b, 0, 0, n);
    }

    // 添字でアクセス
    T operator[](int i) {
        return data[i + n - 1];
    }
};

#define R_MIN ([](long long a, long long b) { return min(a, b); })
#define R_MAX ([](long long a, long long b) { return max(a, b); })
#define R_SUM ([](long long a, long long b) { return a + b; })

#define NORMAL_UPDATE ([](long long a, long long b) { return b; })
#define ADD_UPDATE ([](long long a, long long b) { return a + b; })
#define MINUS_UPDATE ([](long long a, long long b) { return a - b; }
//Seg-Tree-End

//dfs
//vector<int> v[100004];
//bool went[100004];
//void dfs(int x) {
//    went[x] = true;
//    for (int i = 0; i < (int)v[x].size(); i++) {
//        if (!went[v[x][i]]) {
//            dfs(v[x][i]);
//        }
//    }
//}

//number of keta
int keta(int x) {
    int ans = 0;
    while (x) {
        x /= 10;
        ans++;
    }
    return ans;
}

//sum of keta
int ketasum(int x) {
    int ans = 0;
    while (x) {
        ans += x % 10;
        x /= 10;
    }
    return ans;
}

inline int lcm(int x, int y) {
    int ans = x / gcd(x, y) * y;
    return ans;
}
int twobeki(int x) {
    int ans = 0;
    while (1) {
        if (!(x & 1)) {
            ans++;
            x /= 2;
        }
        else {
            break;
        }
    }
    return ans;
}

int kai(int x, int y) {
    int res = 1;
    for (int i = x - y + 1; i <= x; i++) {
        res *= i; res %= mod;
    }
    return res;
}

int comb(int x, int y) {
    if (y > x)return 0;
//    cout<<kai(x, y)<<' '<<modpow(kai(y, y), mod - 2)<<endl;
    return kai(x, y) * modpow(kai(y, y), mod - 2) % mod;
}

int fukugen(vector<int> l) {
    int ans = 0;
    for (int i = 0; i < (int)l.size(); i++) {
        ans *= 10;
        ans += l[i];
    }
    return ans;
}
int nanjyo(int n){
    int ans=0;
    while (n>1) {
        ans++;
        n/=2;
    }
    return ans;
}
bool intersect(pair<pair<int,int>, pair<int,int>>p1 ,pair<pair<int,int>,pair<int,int>>p2){
    int men11=(p2.first.first-p1.first.first)*(p2.second.second-p1.first.second)-(p2.second.first-p1.first.first)*(p2.first.second-p1.first.second);
    int men12=(p2.first.first-p1.second.first)*(p2.second.second-p1.second.second)-(p2.second.first-p1.second.first)*(p2.first.second-p1.second.second);
    int men21=(p1.first.first-p2.first.first)*(p1.second.second-p2.first.second)-(p1.second.first-p2.first.first)*(p1.first.second-p2.first.second);
    int men22=(p1.first.first-p2.second.first)*(p1.second.second-p2.second.second)-(p1.second.first-p2.second.first)*(p1.first.second-p2.second.second);
    
    return ((signbit(men11)) ^ (signbit(men12))) && ((signbit(men21)) ^ (signbit(men22)));
}
template <class T, class U>
inline bool chmax(T& lhs, const U& rhs) {
    if (lhs < rhs) {
        lhs = rhs;
        return 1;
    }
    return 0;
}
template <class T, class U>
inline bool chmin(T& lhs, const U& rhs) {
    if (lhs > rhs) {
        lhs = rhs;
        return 1;
    }
    return 0;
}
void Yes(){
    cout<<"Yes"<<endl;
}
void No(){
    cout<<"No"<<endl;
}
void YES(){
    cout<<"YES"<<endl;
}
void NO(){
    cout<<"NO"<<endl;
}
int yakusuu(int x){
    int ans=0;
    for (int i=1; i*i<=x; i++) {
        if (!(x%i)) {
            if (i*i==x) {
                ans++;
            }
            else ans+=2;
        }
    }
    return ans;
}
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,n) for(int i=1;i<=n;i++)
int neocomb(int x,int y){
    if (!x) {
        return 0;
    }
    int ans=1;
    int waru=1;
    for(int i=x+y-1;i>y;i--){
        ans*=i;
        ans%=mod;
        waru*=x+y-i;
        waru%=mod;
    }
    ans*=modpow(waru, mod-2);
    ans%=mod;
    return ans;
}
#define fin(i) scanf("%lld",&i)
#define fout(i) printf("%lld",&i)
//Library-End



int n,k,ans,hiku;
signed main(){
    cin>>n>>k;

    ans=neocomb(n,n);
	
	if(k<n-1){
		hiku=neocomb(n-k-1,n);
		hiku*=comb(k+1,n);
		hiku%=mod;
	}
	ans+=mod;
	ans-=hiku;
	ans%=mod;
	cout<<ans<<endl;
    
}