Quite recently a creative student Lesha had a lecture on trees. After the lecture Lesha was inspired and came up with the tree of his own which he called a *k*-tree.

A *k*-tree is an infinite rooted tree where:
 -  each vertex has exactly *k* children; -  each edge has some weight; -  if we look at the edges that goes from some vertex to its children (exactly *k* edges), then their weights will equal 1, 2, 3, ..., *k*. 
The picture below shows a part of a 3-tree.

 

 <img class="tex-graphics" src="https://espresso.codeforces.com/61ac5ac17c4d2f220968eff8fb59bb3c8dc77edf.png" style="max-width: 100.0%;max-height: 100.0%;"/>

 

Help Dima find an answer to his question. As the number of ways can be rather large, print it modulo 1000000007 (10<sup class="upper-index">9</sup> + 7).
A single line contains three space-separated integers: *n*, *k* and *d* (1 ≤ *n*, *k* ≤ 100; 1 ≤ *d* ≤ *k*).
#include <bits/stdc++.h>
using namespace std;

int f(int n,int k,int s){
    if(s>n){
        return 0;
    }
    if(s==n){
        return 1;
    }
    int ans=0;
    for(int j=1;j<=k;j++){
        ans+=f(n,k,s+j);
    }
    return ans;
}
int main(){
    int mod=1000000007;
    int n,d,k;
    cin>>n>>k>>d;
    int ans=(f(n,k,0)-f(n,d-1,0))%mod;
    cout<<ans;
}


