codeforces431C

k-Tree

CodeForces - 431C

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 kedges), then their weights will equal 1, 2, 3, ..., k.

The picture below shows a part of a 3-tree.

As soon as Dima, a good friend of Lesha, found out about the tree, he immediately wondered: "How many paths of total weight n (the sum of all weights of the edges in the path) are there, starting from the root of a k-tree and also containing at least one edge of weight at least d?".

Help Dima find an answer to his question. As the number of ways can be rather large, print it modulo 1000000007 (10⁹ + 7).

Input

A single line contains three space-separated integers: n, k and d (1 ≤ n, k ≤ 100; 1 ≤ d ≤ k).

Output

Print a single integer — the answer to the problem modulo 1000000007 (10⁹ + 7).

Examples

Input

3 3 2

Output

3

Input

3 3 3

Output

1

Input

4 3 2

Output

6

Input

4 5 2

Output

7

题意：给出K-Tree定义，每个结点都有恰好K个孩子，这棵树无限增长。每个节点到它K个孩子的K条边的权重刚好是1，2，3...,K（看图应该也看得明白）
现在问有多少条路径，使得从根节点出发到达某个结点，经过的边权重之和恰好为n，并且经过的边至少有一条权重不小于d。

sol:dp应该看得出来，状态也很好构建dp[i][j][0,1]表示到第i层和为j是否有不小于d的边,因为n，k太小，毫无思考的n3dp直接上

#include <bits/stdc++.h>
using namespace std;
typedef int ll;
inline ll read()
{
    ll s=;
    bool f=;
    char ch=' ';
    while(!isdigit(ch))
    {
        f|=(ch=='-'); ch=getchar();
    }
    while(isdigit(ch))
    {
        s=(s<<)+(s<<)+(ch^); ch=getchar();
    }
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<)
    {
        putchar('-'); x=-x;
    }
    if(x<)
    {
        putchar(x+''); return;
    }
    write(x/);
    putchar((x%)+'');
    return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('\n')
const int Mod=,N=;
int n,K,D;
int dp[N][N][];
int main()
{
    int i,j,k;
    R(n); R(K); R(D);
    dp[][][]=;
    for(i=;i<=n;i++)
    {
        for(j=i-;j<n;j++)
        {
            for(k=;j+k<=n&&k<=K;k++)
            {
                dp[i][j+k][]+=1ll*dp[i-][j][]%Mod;
                dp[i][j+k][]-=(dp[i][j+k][]>=Mod)?Mod:;
                if(k>=D)
                {
                    dp[i][j+k][]+=1ll*dp[i-][j][]%Mod;
                    dp[i][j+k][]-=(dp[i][j+k][]>=Mod)?Mod:;
                }
                else
                {
                    dp[i][j+k][]+=1ll*dp[i-][j][]%Mod;
                    dp[i][j+k][]-=(dp[i][j+k][]>=Mod)?Mod:;
                }
            }
        }
    }
    int ans=;
    for(i=;i<=n;i++)
    {
        ans+=dp[i][n][];
        ans-=(ans>=Mod)?Mod:;
    }
    Wl(ans);
    return ;
}
/*
input
3 3 2
output
3

input
3 3 3
output
1

input
4 3 2
output
6

input
4 5 2
output
7
*/