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 <iostream>
#include <vector>
#include <queue>
using namespace std;


int main(){
    int n,k,d;
    cin>>n>>k>>d;
    //n-=d;
    //if (n<=0){cout<<0;return 0 ;}
    queue<vector<int>>q;
    q.push({0,0});
    int sc = 0;
    while(!q.empty()){
        int ln = q.front()[0], is_d = q.front()[1];
        //cout<<ln<<" "<<is_d<<" \n";
        q.pop();
        if (ln>n) continue;
        if (ln == n && is_d){ sc+=1;}
        for(int i= 1; i <= k; i++){
            if (i>=d) q.push({ln+i,1});
            else q.push({ln+i,is_d});
        }

    }
    cout<<sc;


}
