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
#include<bits/stdc++.h>
using namespace std;
using Int = long long;
template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}


template<typename T,T MOD = 1000000007>
struct Mint{
  static constexpr T mod = MOD;
  T v;
  Mint():v(0){}
  Mint(signed v):v(v){}
  Mint(long long t){v=t%MOD;if(v<0) v+=MOD;}

  Mint pow(long long k){
    Mint res(1),tmp(v);
    while(k){
      if(k&1) res*=tmp;
      tmp*=tmp;
      k>>=1;
    }
    return res;
  }

  static Mint add_identity(){return Mint(0);}
  static Mint mul_identity(){return Mint(1);}

  Mint inv(){return pow(MOD-2);}

  Mint& operator+=(Mint a){v+=a.v;if(v>=MOD)v-=MOD;return *this;}
  Mint& operator-=(Mint a){v+=MOD-a.v;if(v>=MOD)v-=MOD;return *this;}
  Mint& operator*=(Mint a){v=1LL*v*a.v%MOD;return *this;}
  Mint& operator/=(Mint a){return (*this)*=a.inv();}

  Mint operator+(Mint a) const{return Mint(v)+=a;};
  Mint operator-(Mint a) const{return Mint(v)-=a;};
  Mint operator*(Mint a) const{return Mint(v)*=a;};
  Mint operator/(Mint a) const{return Mint(v)/=a;};

  Mint operator-() const{return v?Mint(MOD-v):Mint(v);}

  bool operator==(const Mint a)const{return v==a.v;}
  bool operator!=(const Mint a)const{return v!=a.v;}
  bool operator <(const Mint a)const{return v <a.v;}

  // find x s.t. a^x = b
  static T log(T a,T b){
    const T sq=40000;
    unordered_map<T, T> dp;
    dp.reserve(sq);
    Mint res(1);
    for(Int r=0;r<sq;r++){
      if(!dp.count(res.v)) dp[res.v]=r;
      res*=a;
    }
    Mint p=Mint(a).inv().pow(sq);
    res=b;
    for(Int q=0;q<=MOD/sq+1;q++){
      if(dp.count(res.v)){
        T idx=q*sq+dp[res.v];
        if(idx>0) return idx;
      }
      res*=p;
    }
    assert(0);
    return T(-1);
  }

  static Mint comb(long long n,Int k){
    Mint num(1),dom(1);
    for(Int i=0;i<k;i++){
      num*=Mint(n-i);
      dom*=Mint(i+1);
    }
    return num/dom;
  }
};
template<typename T,T MOD> constexpr T Mint<T, MOD>::mod;
template<typename T,T MOD>
ostream& operator<<(ostream &os,Mint<T, MOD> m){os<<m.v;return os;}


template<typename M>
class Enumeration{
private:
  static vector<M> fact,finv,invs;
public:
  static void init(Int n){
    n=min<decltype(M::mod)>(n,M::mod-1);

    Int m=fact.size();
    if(n<m) return;

    fact.resize(n+1,1);
    finv.resize(n+1,1);
    invs.resize(n+1,1);

    if(m==0) m=1;
    for(Int i=m;i<=n;i++) fact[i]=fact[i-1]*M(i);
    finv[n]=M(1)/fact[n];
    for(Int i=n;i>=m;i--) finv[i-1]=finv[i]*M(i);
    for(Int i=m;i<=n;i++) invs[i]=finv[i]*fact[i-1];
  }

  static M Fact(Int n){
    init(n);
    return fact[n];
  }
  static M Finv(Int n){
    init(n);
    return finv[n];
  }
  static M Invs(Int n){
    init(n);
    return invs[n];
  }

  static M C(Int n,Int k){
    if(n<k||k<0) return M(0);
    init(n);
    return fact[n]*finv[n-k]*finv[k];
  }

  static M P(Int n,Int k){
    if(n<k||k<0) return M(0);
    init(n);
    return fact[n]*finv[n-k];
  }

  static M H(Int n,Int k){
    if(n<0||k<0) return M(0);
    if(!n&&!k) return M(1);
    init(n+k-1);
    return C(n+k-1,k);
  }

  static M S(Int n,Int k){
    M res;
    init(k);
    for(Int i=1;i<=k;i++){
      M tmp=C(k,i)*M(i).pow(n);
      if((k-i)&1) res-=tmp;
      else res+=tmp;
    }
    return res*=finv[k];
  }

  static vector<vector<M> > D(Int n,Int m){
    vector<vector<M> > dp(n+1,vector<M>(m+1,0));
    dp[0][0]=M(1);
    for(Int i=0;i<=n;i++){
      for(Int j=1;j<=m;j++){
        if(i-j>=0) dp[i][j]=dp[i][j-1]+dp[i-j][j];
        else dp[i][j]=dp[i][j-1];
      }
    }
    return dp;
  }

  static M B(Int n,Int k){
    if(n==0) return M(1);
    k=min(k,n);
    init(k);
    vector<M> dp(k+1);
    dp[0]=M(1);
    for(Int i=1;i<=k;i++)
      dp[i]=dp[i-1]+((i&1)?-finv[i]:finv[i]);
    M res;
    for(Int i=1;i<=k;i++)
      res+=M(i).pow(n)*finv[i]*dp[k-i];
    return res;
  }

  static M montmort(Int n){
    M res;
    init(n);
    for(Int k=2;k<=n;k++){
      if(k&1) res-=finv[k];
      else res+=finv[k];
    }
    return res*=fact[n];
  }

  static M LagrangePolynomial(vector<M> &y,M t){
    Int n=y.size()-1;
    if(t.v<=n) return y[t.v];
    init(n+1);
    vector<M> dp(n+1,1),pd(n+1,1);
    for(Int i=0;i<n;i++) dp[i+1]=dp[i]*(t-M(i));
    for(Int i=n;i>0;i--) pd[i-1]=pd[i]*(t-M(i));
    M res{0};
    for(Int i=0;i<=n;i++){
      M tmp=y[i]*dp[i]*pd[i]*finv[i]*finv[n-i];
      if((n-i)&1) res-=tmp;
      else res+=tmp;
    }
    return res;
  }
};
template<typename M>
vector<M> Enumeration<M>::fact = vector<M>();
template<typename M>
vector<M> Enumeration<M>::finv = vector<M>();
template<typename M>
vector<M> Enumeration<M>::invs = vector<M>();

//INSERT ABOVE HERE
signed main(){
  using M = Mint<Int>;
  using E = Enumeration<M>;

  Int n,m,l,r;
  cin>>n>>m>>l>>r;

  const Int MAX = 3e5+100;
  E::init(MAX*2);

  vector<M> ps(MAX);
  for(Int k=1;k<MAX;k++){
    for(Int i=0;i<=n-m;i++){
      if(k*m+k*i>=MAX) break;
      if(~i&1)
        ps[k*m+k*i]+=E::C(n-m,i);
      else
        ps[k*m+k*i]-=E::C(n-m,i);
    }
  }

  vector<M> qs(MAX);
  for(Int i=0;i<MAX;i++) qs[i]=E::C(i+n,n);
  for(Int i=MAX-1;i>=m;i--) qs[i]-=qs[i-m];

  auto calc=
    [&](Int d){
      M res{0};
      for(Int i=0;i<=d;i++){
        Int j=d-i;
        res+=ps[i]*qs[j];
      }
      return res;
    };
  M res=calc(r)-calc(l-1);

  M ans{0};
  for(Int i=l;i<=r;i++) ans+=E::H(n,i);
  ans-=res*E::C(n,m);
  cout<<ans<<endl;
  return 0;
}
