#include <bits/stdc++.h>

using namespace std;

#define int long long

const int modulo = 1e9 + 7;

int k,s,t;
int fact[5000005],invfact[5000005];

int C(int x,int y)
{
    if (x < y)
        return 0;
    if (y < 0)
        return 0;
    return fact[x] * invfact[y] % modulo * invfact[x - y] % modulo;
}

int f(int x,int y)
{
    if (x < y)
        return 0;
    if (x == 0 and y == 0)
        return 1;
    if (y == 0)
        return 0;
    ///in cate moduri pot distribui x in y numere pozitive
    ///in cate moduri pot pune y - 1 bare in x - 1 pozitii cu oricare doua bare diferite
    ///C(x - 1,y - 1)
    return C(x - 1,y - 1);
}

int lgpow(int x,int y)
{
    int z = 1;
    while (y != 0)
    {
        if (y % 2 == 1)
            z = z * x % modulo;
        x = x * x % modulo;
        y /= 2;
    }
    return z;
}

void prec()
{
    fact[0] = invfact[0] = 1;
    for (int i = 1; i <= 5e6; i++)
        fact[i] = i * fact[i - 1] % modulo;
    invfact[(int)5e6] = lgpow(fact[(int)5e6],modulo - 2);
    for (int i = 5e6 - 1; i >= 1; i--)
        invfact[i] = invfact[i + 1] * (i + 1) % modulo;
}

signed main()
{
    prec();
    cin >> k >> s >> t;
    int d = k / t,r = k % t;
    int ans = 0;
    for (int x = 0; x <= s / (d + 1); x++)
    {
        if ((s - (d + 1) * x) % d == 0)
            ans = (ans + f(x,r) * f((s - (d + 1) * x) / d,t - r)) % modulo;
    }
    cout << ans;
    return 0;
}

/**
int d = k / t,r = k % t
(d + 1)(a[1] + a[2] + ... + a[r]) + d(a[r + 1] + ... + a[k])
fie x = a[1] + ... + a[r]
pentru fiecare x posibil, avem:
0 daca (s - (d + 1)x) % d != 0
f(x,r) * f((s - (d + 1)x) / d,k - r) altfel
si avem suma din asta
**/