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;
//#pragma GCC optimize("Ofast")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
 
#define ms(s, n) memset(s, n, sizeof(s))
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define FORd(i, a, b) for (int i = (a) - 1; i >= (b); --i)
#define FORall(it, a) for (__typeof((a).begin()) it = (a).begin(); it != (a).end(); it++)
#define sz(a) int((a).size())
#define present(t, x) (t.find(x) != t.end())
#define all(a) (a).begin(), (a).end()
#define uni(a) (a).erase(unique(all(a)), (a).end())
#define pb push_back
#define pf push_front
#define mp make_pair
#define fi first
#define se second
#define prec(n) fixed<<setprecision(n)
#define bit(n, i) (((n) >> (i)) & 1)
#define bitcount(n) __builtin_popcountll(n)
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pi;
typedef vector<int> vi;
typedef vector<pi> vii;
const int MOD = (int) 1e9 + 7;
const int FFTMOD = 119 << 23 | 1;
const int INF = (int) 1e9 + 23111992;
const ll LINF = (ll) 1e18 + 23111992;
const ld PI = acos((ld) -1);
const ld EPS = 1e-9;
inline ll gcd(ll a, ll b) {ll r; while (b) {r = a % b; a = b; b = r;} return a;}
inline ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
inline ll fpow(ll n, ll k, int p = MOD) {ll r = 1; for (; k; k >>= 1) {if (k & 1) r = r * n % p; n = n * n % p;} return r;}
template<class T> inline int chkmin(T& a, const T& val) {return val < a ? a = val, 1 : 0;}
template<class T> inline int chkmax(T& a, const T& val) {return a < val ? a = val, 1 : 0;}
inline ull isqrt(ull k) {ull r = sqrt(k) + 1; while (r * r > k) r--; return r;}
inline ll icbrt(ll k) {ll r = cbrt(k) + 1; while (r * r * r > k) r--; return r;}
inline void addmod(int& a, int val, int p = MOD) {if ((a = (a + val)) >= p) a -= p;}
inline void submod(int& a, int val, int p = MOD) {if ((a = (a - val)) < 0) a += p;}
inline int mult(int a, int b, int p = MOD) {return (ll) a * b % p;}
inline int inv(int a, int p = MOD) {return fpow(a, p - 2, p);}
inline int sign(ld x) {return x < -EPS ? -1 : x > +EPS;}
inline int sign(ld x, ld y) {return sign(x - y);}
mt19937 mt(chrono::high_resolution_clock::now().time_since_epoch().count());
inline int mrand() {return abs((int) mt());}
#define db(x) cerr << "[" << #x << ": " << (x) << "] ";
#define endln cerr << "\n";

void solve() {
    int n, m, l, r; cin >> n >> m >> l >> r;
    vi fac(n + r + 1 << 1), ifac(n + r + 1 << 1);
    fac[0] = 1; FOR(i, 1, sz(fac)) fac[i] = mult(fac[i - 1], i);
    FOR(i, 0, sz(fac)) ifac[i] = inv(fac[i]);
    auto binom = [&] (int n, int k) {
        if (n < k) return 0;
        return mult(fac[n], mult(ifac[k], ifac[n - k]));
    };
    vi f(r + 1);
    FOR(i, 0, n - m + 1) if (m + i <= r) {
        f[m + i] = binom(n - m, i);
        if (i & 1) {
            f[m + i] = (MOD - f[m + i]) % MOD;
        }
    }
    vector<vi> g(r + 1);
    FOR(i, 1, r + 1) {
        for (int j = 0; j <= r; j += i) {
            g[j].pb(i);
        }
    }
    vi h(r + 1);
    FOR(i, 1, r + 1) {
        for (int x : g[i]) {
            addmod(h[i], f[x]);
        }
    }
    auto func = [&] (int t) {
        int res = 0;
        FOR(i, 0, t + 1) {
            int coef = binom(n + i, i);
            if (m <= i) {
                submod(coef, binom(n + i - m, i - m));
            }
            addmod(res, mult(coef, h[t - i]));
        }
        return mult(res, binom(n, m));
    };
    int res = binom(r + n, n);
    if (l) submod(res, binom(l - 1 + n, n));
    submod(res, func(r));
    if (l) addmod(res, func(l - 1));
    cout << res << "\n";
}

int main(int argc, char* argv[]) {
    ios_base::sync_with_stdio(0), cin.tie(0);
    if (argc > 1) {
        assert(freopen(argv[1], "r", stdin));
    }
    if (argc > 2) {
        assert(freopen(argv[2], "wb", stdout));
    }
    solve();
    cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
    return 0;
}
