C. k-Tree

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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 equal1, 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

解题说明：题意是给出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。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。做法是采用动态规划

dp[i][0]:表示权值和为i中不包含权值>=d的边。

dp[i][j]:表示权值和为i中包含权值>=d的边。

dp[i][0]+=dp[i-j][0] (j<d)

dp[i][1]+=dp[i-j][0] (j>=d)

dp[i][1]+=dp[i-j][1];

#include<cstdio>
#include <cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#define MAX_N (100)
#define MOD (1000000007)
using namespace std;

int dp[2][MAX_N+1];

void build_dp(int n, int k, int d)
{
int j,m;
dp[0][0] = 1;
for ( j=1; j <= n; ++j )
{
for ( m=1; m <= k; ++m )
{
dp[0][j] += j-m >= 0 ? dp[0][j-m] : 0, dp[0][j] %= MOD;
}
}
for ( j=1; j <= n; ++j )
{
for ( m=1; m < d; ++m )
{
dp[1][j] += j-m >= 0 ? dp[1][j-m] : 0, dp[1][j] %= MOD;
}
for ( m=d; m <= k; ++m )
{
dp[1][j] += j-m >= 0 ? dp[0][j-m] : 0, dp[1][j] %= MOD;
}
}
}

int main()
{
int n,k,d;
scanf("%d %d %d", &n, &k, &d);
build_dp(n, k, d);
printf("%d\n", dp[1]
);
return 0;
}