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>
#include <unistd.h>
 
using namespace std;
 
#define DEBUG(x) cerr<<#x<<": "<<x<<endl;
#define DEBUG_VEC(v) cerr<<#v<<":";for(int i=0;i<v.size();i++) cerr<<" "<<v[i]; cerr<<endl
#define DEBUG_MAT(v) cerr<<#v<<endl;for(int i=0;i<v.size();i++){for(int j=0;j<v[i].size();j++) {cerr<<v[i][j]<<" ";}cerr<<endl;}
 
 
typedef long long ll;

#define vi vector<int>
#define vl vector<ll>
#define vii vector< vector<int> >
#define vll vector< vector<ll> >
#define vs vector<string>
#define pii pair<int,int>
#define pis pair<int,string>
#define psi pair<string,int>
#define pll pair<ll,ll>
template<class S, class T> pair<S, T> operator+(const pair<S, T> &s, const pair<S, T> &t) { return pair<S, T>(s.first + t.first, s.second + t.second); }
template<class S, class T> pair<S, T> operator-(const pair<S, T> &s, const pair<S, T> &t) { return pair<S, T>(s.first - t.first, s.second - t.second); }
template<class S, class T> ostream& operator<<(ostream& os, pair<S, T> p) { os << "(" << p.first << ", " << p.second << ")"; return os; }
#define X first
#define Y second
#define rep(i,n) for(int i=0;i<(n);i++)
#define rep1(i,n) for(int i=1;i<=(n);i++)
#define rrep(i,n) for(int i=(n)-1;i>=0;i--)
#define rrep1(i,n) for(int i=(n);i>0;i--)
#define REP(i,a,b) for(int i=a;i<b;i++)
#define in(x, a, b) (a <= x && x < b)
#define all(c) c.begin(),c.end()
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a = b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a = b; return 1; } return 0; }
#define UNIQUE(v) v.erase(std::unique(v.begin(), v.end()), v.end());
const ll inf = 1000000001;
const ll INF = (ll)1e18 + 1;
const ll MOD = 1000000007;
//const ll MOD = 998244353;
const double pi = 3.14159265358979323846;
#define Sp(p) cout<<setprecision(15)<< fixed<<p<<endl;
int dx[4] = { -1,0, 1,0 }, dy[4] = { 0,1,0,-1 };
int dx2[8] = { 1,1,0,-1,-1,-1,0,1 }, dy2[8] = { 0,1,1,1,0,-1,-1,-1 };
#define fio() cin.tie(0); ios::sync_with_stdio(false);
// #define mp make_pair
//#define endl '\n'


const int MAXN = 1000010;

vl fact(MAXN);
vl rfact(MAXN);

ll mod_pow(ll x, ll p, ll M = MOD) {
  ll a = 1;
  while (p) {
    if (p % 2)
      a = a*x%M;
    x = x*x%M;
    p /= 2;
  }
  return a;
}

ll mod_inverse(ll a, ll M = MOD) {
  return mod_pow(a, M - 2, M);
}

void set_fact(ll n, ll M = MOD) {
  fact[0] = fact[1] = rfact[0] = rfact[1] = 1;
  for (ll i = 2; i <= n; i++) {
    fact[i] = i * fact[i - 1] % M;
    // rfact[i] = mod_inverse(fact[i], M);
  }
}

//http://drken1215.hatenablog.com/entry/2018/06/08/210000
//nが大きくfactが計算できないときのほかの計算方法について書いてある
ll nCr(ll n, ll r, ll M = MOD) {
  if (r > n) return 0;
  assert(fact[2] == 2);
  ll ret = fact[n];
  if (rfact[r] == 0) {
    rfact[r] = mod_inverse(fact[r], M);
  }
  ret = (ret*rfact[r]) % M;
  if (rfact[n - r] == 0) {
    rfact[n - r] = mod_inverse(fact[n - r], M);
  }
  ret = (ret*rfact[n - r]) % M;
  return ret;
}

ll n, m, l, r;



ll f(int a, int b) {
  ll ans = 0;
  for (int out = 0; out <= n - m; out++) {
    ll rr = r - a * m;
    rr = rr - out * (b + 1);
    if (rr < 0) break;
    if (out % 2 == 0) {
      ans += nCr(n + rr, n) * nCr(n, m) % MOD * nCr(n - m, out) % MOD;
    }
    else {
      ans -= nCr(n + rr, n) * nCr(n, m) % MOD * nCr(n - m, out) % MOD;
    }
    ans %= MOD;
  }

  for (int out = 0; out <= n - m; out++) {
    ll lll = (l - 1) - a * m;
    lll = lll - out * (b + 1);
    if (lll < 0) break;
    if (out % 2 == 0) {
      ans -= nCr(n + lll, n) * nCr(n, m) % MOD * nCr(n - m, out) % MOD;
    }
    else {
      ans += nCr(n + lll, n) * nCr(n, m) % MOD * nCr(n - m, out) % MOD;
    }
    ans %= MOD;
  }
  return ans;
}


int main() {
  cin >> n >> m >> l >> r;
  set_fact(n + r);
  
  ll zen = nCr(n + r, n) - nCr(n + (l - 1), n);
  zen %= MOD;
  
  rep1 (x, r + 1) {
    if ((ll)x * m > r) break;

    ll sub = f(x, x - 1) - f(x + 1, x - 1);
    zen -= sub;
    zen %= MOD;
  }
  zen = (zen + MOD) % MOD;
  cout << zen << endl;
    
}