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;

template<int MAX>
struct Combination {
    vector<long long> _inv, _fac, _ifac;
    const long long MOD;
    Combination(long long mod) : _inv(MAX+1), _fac(MAX+1), _ifac(MAX+1), MOD(mod) { init(); }
    inline void init(long long mod = 0) { init_inv(), init_fac(); }

    inline void init_inv() {
        _inv[0] = 0;
        for (int i = 1; i <= MAX; i++) {
            if (i == 1) _inv[i] = 1;
            else {
                _inv[i] = (MOD - (MOD / i) * _inv[MOD % i]) % MOD;
                if (_inv[i] < 0) _inv[i] += MOD;
            }
        }
    }
    inline void init_fac() {
        _fac[0] = _ifac[0] = 1;
        for (int i = 1; i <= MAX; i++) {
            _fac[i] = _fac[i-1] * i % MOD;
            _ifac[i] = _ifac[i-1] * _inv[i] % MOD;
        }
    }

    inline long long modpow(long long n, long long r) {
        n %= MOD, r %= (MOD-1);
        if (!n) return 0;
        long long ret = 1; long long tmp = n;
        while (r != 0) {
            if (r % 2) ret *= tmp;
            tmp *= tmp; tmp %= MOD; ret %= MOD;
            r /= 2;
        }
        return ret;
    }
    inline const long long invb(int i) { return (i <= MAX) ? _inv[i] : this->modpow(i, MOD-2); }
    inline const long long inv(int i) { return _inv[i]; }
    inline const long long fac(int i) { return _fac[i]; }
    inline const long long ifac(int i) { return _ifac[i]; }

    const long long operator()(int n, int r) {
        if (n < r) return 0;
        return fac(n) * ifac(r) % MOD * ifac(n-r) % MOD;
    }
};

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    constexpr int MAX = 600000;
    constexpr int mod = 1000000007;
    Combination<MAX> C(mod);
    int n, m, l, r; cin >> n >> m >> l >> r;
    vector<long long> x(r+1, 0);
    vector<long long> y(r+1, 0);
    for (int a = 1; a <= r; a++) {
        for (int i = 0; i <= n-m; i++) {
            if ((long long) (m+i) * a > r) break;
            if (i & 1) x[(m+i)*a] -= C(n-m, i);
            else x[(m+i)*a] += C(n-m, i);
            x[(m+i)*a] = (x[(m+i)*a] + mod) % mod;
        }
    }
    for (int i = r; i >= m; i--) {
        x[i] = (x[i] - x[i-m] + mod) % mod;
    }
    for (int i = 0; i <= r; i++) y[i] = C(n+i, i);
    auto calc = [&](int l) {
        long long ret = 0;
        for (int i = 0; i <= l; i++) {
            ret += x[i] * y[l-i] % mod;
        }
        return ret % mod;
    };
    long long sub = (calc(r) - calc(l-1) + mod) % mod * C(n, m) % mod;
    long long all = 0;
    for (int i = l; i <= r; i++) all += C(i+n-1, i);
    cout << (all - sub + mod) % mod << "\n";
    return 0;
}