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 <iostream>
#include <string>
#include <queue>
#include <stack>
#include <algorithm>
#include <list>
#include <vector>
#include <complex>
#include <utility>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <climits>
#include <bitset>
#include <ctime>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <cassert>
#include <cstddef>
#include <iomanip>
#include <numeric>
#include <tuple>
#include <sstream>
#include <fstream>
#include <functional>

using namespace std;
#define REP(i, n) for (int (i) = 0; (i) < (n); (i)++)
#define FOR(i, a, b) for (int (i) = (a); (i) < (b); (i)++)
#define RREP(i, a) for(int (i) = (a) - 1; (i) >= 0; (i)--)
#define FORR(i, a, b) for(int (i) = (a) - 1; (i) >= (b); (i)--)
#define DEBUG(C) cerr << #C << " = " << C << endl;
using LL = long long;
using VI = vector<int>;
using VVI = vector<VI>;
using VL = vector<LL>;
using VVL = vector<VL>;
using VD = vector<double>;
using VVD = vector<VD>;
using PII = pair<int, int>;
using PDD = pair<double, double>;
using PLL = pair<LL, LL>;
using VPII = vector<PII>;
template<typename T> using VT = vector<T>;
#define ALL(a) begin((a)), end((a))
#define RALL(a) rbegin((a)), rend((a))
#define SORT(a) sort(ALL((a)))
#define RSORT(a) sort(RALL((a)))
#define REVERSE(a) reverse(ALL((a)))
#define MP make_pair
#define FORE(a, b) for (auto &&a : (b))
#define EB emplace_back
#define GREATER(T) T, VT<T> greater<T>
template<typename T> inline bool chmax(T &a, T b){if (a < b){a = b;return true;}return false;}
template<typename T> inline bool chmin(T &a, T b){if (a > b){a = b;return true;}return false;}

const int INF = 1e9;
const int MOD = INF + 7;
const LL LLINF = 1e18;

// 剰余計算,MOD計算,順列・組み合わせ,Permutation・Combination

template<int mod>
class mod_calc {
private:
    const int defnum = -1;
public:
    mod_calc() { }
    const inline int add(int a, const int b) const {
        a += b;
        while (a < 0) a += mod;
        while (a >= mod) a -= mod;
        return a;
    }
    const inline int mul(const int a, const int b) const {
        return (long long)a * b % mod;
    }
    const inline void add_ref(int &a, const int b) const {
        a = this->add(a, b);
    }
    const inline void mul_ref(int &a, const int b) const {
        a = this->mul(a, b);
    }
    const inline void add_ref(long long &a, const int b) const {
        a = this->add(a, b);
    }
    const inline void mul_ref(long long &a, const int b) const {
        a = this->mul(a, b);
    }
    const inline int pow(int a, int p) const {
        int res = 1;
        while (p > 0) {
            if (p & 1) res = this->mul(res, a);
            a = this->mul(a, a);
            p >>= 1;
        }
        return res;
    }
    const inline int inverse(int a) const {
        return this->pow(a, mod - 2);
    }
};

const int MAX = 5e5 + 10;
mod_calc<MOD> mc;
int memo_fact[MAX], memo_inv_fact[MAX];

inline int fact(int n) {
    if (memo_fact[n] != -1) return memo_fact[n];
    return memo_fact[n] = (n == 0 ? 1 : mc.mul(n, fact(n - 1)));
}

inline int inv_fact(int n) {
    if (memo_inv_fact[n] != -1) return memo_inv_fact[n];
    return memo_inv_fact[n] = mc.inverse(fact(n));
}

inline long long perm(int n, int r) {
    return mc.mul(fact(n), inv_fact(n - r));
}

inline long long comb(int n, int r) {
    return mc.mul(mc.mul(fact(n), inv_fact(r)), inv_fact(n - r));
}

// M個の物をN人で分ける
inline long long comb_with_repetition(int M, int N) {
    return comb(M + N - 1, N - 1);
}

//最初に呼んで♪ O(MAX)
inline void init() {
    memset(memo_fact, -1, sizeof(memo_fact));
    memset(memo_inv_fact, -1, sizeof(memo_inv_fact));
    for (int i = 0; i < MAX; i++) {
        fact(i);
        inv_fact(i);
    }
}

int main(void) {
    init();
    int N, K;
    cin >> N >> K;
    if (N <= K + 1) {
        cout << comb_with_repetition(N, N) << endl;
        return 0;
    }

    // 0 個の部屋が K 個以下
    LL ans = comb_with_repetition(N, N);
    FOR(zero_cnt, K + 1, N) {
        LL tmp = comb(N - 1, N - zero_cnt - 1);
        const auto tmp2 = mc.mul(fact(N), mc.mul(inv_fact(N - zero_cnt), inv_fact(zero_cnt)));
        tmp = mc.mul(tmp, tmp2);
        ans = mc.add(ans, -tmp);
    }

    cout << ans << endl;
}