Problem: 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.

Examples

Input

4 2 3 7


Output

105


Input

2 1 4 8


Output

3


Input

141592 6535 89793 238462


Output

933832916

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define rep1(i,n) for(int i=1;i<=(int)(n);i++)
#define all(c) c.begin(),c.end()
#define pb push_back
#define fs first
#define sc second
#define chmin(x,y) x=min(x,y)
#define chmax(x,y) x=max(x,y)
using namespace std;
template<class S,class T> ostream& operator<<(ostream& o,const pair<S,T> &p){
	return o<<"("<<p.fs<<","<<p.sc<<")";
}
template<class T> ostream& operator<<(ostream& o,const vector<T> &vc){
	o<<"{";
	for(const T& v:vc) o<<v<<",";
	o<<"}";
	return o;
}
using ll = long long;
template<class T> using V = vector<T>;
template<class T> using VV = vector<vector<T>>;
constexpr ll TEN(int n) { return (n == 0) ? 1 : 10 * TEN(n-1); }

#ifdef LOCAL
#define show(x) cerr << "LINE" << __LINE__ << " : " << #x << " = " << (x) << endl
#else
#define show(x) true
#endif

template<unsigned int mod_>
struct ModInt{
	using uint = unsigned int;
	using ll = long long;
	using ull = unsigned long long;

	constexpr static uint mod = mod_;

	uint v;
	ModInt():v(0){}
	ModInt(ll _v):v(normS(_v%mod+mod)){}
	explicit operator bool() const {return v!=0;}
	static uint normS(const uint &x){return (x<mod)?x:x-mod;}		// [0 , 2*mod-1] -> [0 , mod-1]
	static ModInt make(const uint &x){ModInt m; m.v=x; return m;}
	ModInt operator+(const ModInt& b) const { return make(normS(v+b.v));}
	ModInt operator-(const ModInt& b) const { return make(normS(v+mod-b.v));}
	ModInt operator-() const { return make(normS(mod-v)); }
	ModInt operator*(const ModInt& b) const { return make((ull)v*b.v%mod);}
	ModInt operator/(const ModInt& b) const { return *this*b.inv();}
	ModInt& operator+=(const ModInt& b){ return *this=*this+b;}
	ModInt& operator-=(const ModInt& b){ return *this=*this-b;}
	ModInt& operator*=(const ModInt& b){ return *this=*this*b;}
	ModInt& operator/=(const ModInt& b){ return *this=*this/b;}
	ModInt& operator++(int){ return *this=*this+1;}
	ModInt& operator--(int){ return *this=*this-1;}
	ll extgcd(ll a,ll b,ll &x,ll &y) const{
		ll p[]={a,1,0},q[]={b,0,1};
		while(*q){
			ll t=*p/ *q;
			rep(i,3) swap(p[i]-=t*q[i],q[i]);
		}
		if(p[0]<0) rep(i,3) p[i]=-p[i];
		x=p[1],y=p[2];
		return p[0];
	}
	ModInt inv() const {
		ll x,y;
		extgcd(v,mod,x,y);
		return make(normS(x+mod));
	}
	ModInt pow(ll p) const {
		if(p<0) return inv().pow(-p);
		ModInt a = 1;
		ModInt x = *this;
		while(p){
			if(p&1) a *= x;
			x *= x;
			p >>= 1;
		}
		return a;
	}
	bool operator==(const ModInt& b) const { return v==b.v;}
	bool operator!=(const ModInt& b) const { return v!=b.v;}
	friend istream& operator>>(istream &o,ModInt& x){
		ll tmp;
		o>>tmp;
		x=ModInt(tmp);
		return o;
	}
	friend ostream& operator<<(ostream &o,const ModInt& x){ return o<<x.v;}
};
using mint = ModInt<1000000007>;

V<mint> fact,ifact;
mint Choose(int a,int b){
	if(b<0 || a<b) return 0;
	return fact[a] * ifact[b] * ifact[a-b];
}
void InitFact(int N){
	fact.resize(N);
	ifact.resize(N);
	fact[0] = 1;
	rep1(i,N-1) fact[i] = fact[i-1] * i;
	ifact[N-1] = fact[N-1].inv();
	for(int i=N-2;i>=0;i--) ifact[i] = ifact[i+1] * (i+1);
}

mint solve(int N,int M,int S){
	int L = N - M, R = M;
	mint ans = 0;
	for(int x=1;x<=S;x++){
		mint pans = ans;
		for(int y : {x,x+1}){
			ll SS = S - (ll)y * R;
			if(SS < 0) continue;
			mint tmp = 0;
			rep(k,L+1){
				ll SSS = SS - (ll)k * x;
				if(SSS < 0) break;
				if(k%2 == 0) tmp += Choose(L,k) * Choose(SSS+N,N);
				else tmp -= Choose(L,k) * Choose(SSS+N,N);
			}
			if(y == x) ans += tmp;
			else ans -= tmp;
		}
		show(x);
		show(ans-pans);
	}
	show(ans);
	ans *= Choose(N,L);
	ans = Choose(S+N,N) - ans;
	return ans;
}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);		//DON'T USE scanf/printf/puts !!
	cout << fixed << setprecision(20);
	InitFact(1000000);
	int N,M,L,R;
	cin >> N >> M >> L >> R;
	cout << solve(N,M,R) - solve(N,M,L-1) << endl;
}