Problem:
Score: 1000 points

Problem Statement

Find the number of sequences of N non-negative integers A_1, A_2, ..., A_N that satisfy the following conditions:

L \leq A_1 + A_2 + ... + A_N \leq R

When the N elements are sorted in non-increasing order, the M-th and (M+1)-th elements are equal.

Since the answer can be enormous, print it modulo 10^9+7.

Constraints

All values in input are integers.

1 \leq M < N \leq 3 \times 10^5

1 \leq L \leq R \leq 3 \times 10^5

Input

Input is given from Standard Input in the following format:

N M L R

Output

Print the number of sequences of N non-negative integers, modulo 10^9+7.

Sample Input 1

4 2 3 7

Sample Output 1

105

The sequences of non-negative integers that satisfy the conditions are:

\begin{eqnarray} &(1, 1, 1, 0), (1, 1, 1, 1), (2, 1, 1, 0), (2, 1, 1, 1), (2, 2, 2, 0), (2, 2, 2, 1), \\ &(3, 0, 0, 0), (3, 1, 1, 0), (3, 1, 1, 1), (3, 2, 2, 0), (4, 0, 0, 0), (4, 1, 1, 0), \\ &(4, 1, 1, 1), (5, 0, 0, 0), (5, 1, 1, 0), (6, 0, 0, 0), (7, 0, 0, 0)\end{eqnarray}

and their permutations, for a total of 105 sequences.

Sample Input 2

2 1 4 8

Sample Output 2

3

The three sequences that satisfy the conditions are (2, 2), (3, 3), and (4, 4).

Sample Input 3

141592 6535 89793 238462

Sample Output 3

933832916

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
//#define cerr if (false) cerr
#define db(x) cerr << #x << "=" << x << endl
#define db2(x, y) cerr << #x << "=" << x << "," << #y << "=" << y << endl
#define db3(x, y, z) cerr << #x << "=" << x << "," << #y << "=" << y << "," << #z << "=" << z << endl
#define dbv(v) cerr << #v << "="; for (auto _x : v) cerr << _x << ", "; cerr << endl
#define dba(a, n) cerr << #a << "="; for (int _i = 0; _i < (n); ++_i) cerr << a[_i] << ", "; cerr << endl
template <typename A, typename B>
ostream& operator<<(ostream& os, const pair<A, B>& x) {
	return os << "(" << x.first << "," << x.second << ")";
}
typedef long long ll;
typedef long double ld;
const int MOD = 1000000007; // 2^23 * 119 + 1
struct Mint {
	int val;
	Mint() { val = 0; }
	Mint(const ll& x) {
		val = (-MOD <= x && x < MOD) ? x : x % MOD;
		if (val < 0) val += MOD;
	}
	template <typename U>
	explicit operator U() const { return (U)val; }
	friend bool operator==(const Mint& a, const Mint& b) { return a.val == b.val; }
	friend bool operator!=(const Mint& a, const Mint& b) { return !(a == b); }
	friend bool operator<(const Mint& a, const Mint& b) { return a.val < b.val; }
	Mint& operator+=(const Mint& m) { if ((val += m.val) >= MOD) val -= MOD; return *this; }
	Mint& operator-=(const Mint& m) { if ((val -= m.val) < 0) val += MOD; return *this; }
	Mint& operator*=(const Mint& m) { val = (ll)val * m.val % MOD; return *this; }
	friend Mint modex(Mint a, ll p) {
		assert(p >= 0);
		Mint ans = 1;
		for (; p; p >>= 1, a *= a) if (p & 1) ans *= a;
		return ans;
	}
	Mint& operator/=(const Mint& m) { return *this *= modex(m, MOD - 2); }
	Mint& operator++() { return *this += 1; }
	Mint& operator--() { return *this -= 1; }
	Mint operator++(int) { Mint result(*this); *this += 1; return result; }
	Mint operator--(int) { Mint result(*this); *this -= 1; return result; }
	Mint operator-() const { return Mint(-val); }
	friend Mint operator+(Mint a, const Mint& b) { return a += b; }
	friend Mint operator-(Mint a, const Mint& b) { return a -= b; }
	friend Mint operator*(Mint a, const Mint& b) { return a *= b; }
	friend Mint operator/(Mint a, const Mint& b) { return a /= b; }
	friend ostream& operator<<(ostream& os, const Mint& x) { return os << x.val; }
};
const int MAXN = 600005;
Mint fac[MAXN], invfac[MAXN];
Mint binom(ll a, ll b) {
	if (b < 0) return 0;
	if (b > a) return 0;
	return fac[a] * invfac[b] * invfac[a - b];
}
int main() {
	// Mint version.
	fac[0] = 1;
	for (int i = 1; i < MAXN; ++i)
		fac[i] = fac[i - 1] * i;
	invfac[MAXN - 1] = 1 / fac[MAXN - 1];
	for (int i = MAXN - 2; i >= 0; --i)
		invfac[i] = invfac[i + 1] * (i + 1);
	int N, M, L, R;
	scanf("%d%d%d%d", &N, &M, &L, &R);
	vector<Mint> sub(R + 1), denom(R + 1);
	Mint w = binom(N, M);
	for (int k = 0; k <= R; ++k) {
		for (int i = 0; i <= N - M; ++i) {
			ll p = (ll)M * k + (ll)i * k;
			if (p > R) break;
			Mint coeff = binom(N - M, i);
			if (i & 1) coeff = -coeff;
			sub[p] += coeff * w;
		}
	}
	for (int i = R; i >= M; --i) {
		sub[i] = sub[i - M] - sub[i];
	}
	sub[0]++;
	sub.resize(R + 1);
	Mint ans = 0;
	for (int i = 0; i <= R; ++i) {
		ans += binom(N + i, i) * sub[R - i];
	}
	for (int i = 0; i <= L - 1; ++i) {
		ans -= binom(N + i, i) * sub[L - 1 - i];
	}
	printf("%d\n", ans.val);
}