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 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

題意：

給定一個k叉樹，且是無限有根樹，每個節點有k個孩子，從左向右依次編號為1,2,3,...k，每條邊有權值，節點連向i號孩子的邊的權值為i，問：從根節點開始向下走，有多少條路徑滿足要求：該路徑權值為ｎ，且路徑上至少有一條邊的權值大於等於ｄ。

題解：

直接求不好求，比較複雜，不過我們可以求出來所以權值為ｎ的路徑，然後再減去不符合要求的路徑：該路徑上所有的邊的權值都小於ｄ。

用dp做。

dp[i]:權值為i的路徑的條數。

dp2[i]:權值為i的，且路徑上的權值都小於d的路徑條數。

答案為:dp[n]-dp2[n]

/****************
*PID:431c div2
*Auth:Jonariguez
*****************
dp[i]:權值為i的路徑的條數
dp2[i]:權值為i，且路徑上的權值都小於d的路徑條數
答案：dp[n]-dp2[n]
*/
#define lson k*2,l,m
#define rson k*2+1,m+1,r
#define rep(i,s,e) for(i=(s);i<=(e);i++)
#define For(j,s,e) For(j=(s);j<(e);j++)
#define sc(x) scanf("%d",&x)
#define In(x) scanf("%I64d",&x)
#define pf(x) printf("%d",x)
#define pfn(x) printf("%d\n",(x))
#define Pf(x) printf("%I64d",(x))
#define Pfn(x) printf("%I64d\n",(x))
#define Pc printf(" ")
#define PY puts("YES")
#define PN puts("NO")
#include <stdio.h>
#include <string.h>
#include <string>
#include <math.h>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
typedef int Ll;
Ll quick_pow(Ll a,Ll b,Ll MOD){a%=MOD;Ll res=1;while(b){if(b&1)res=(res*a)%MOD;b/=2;a=(a*a)%MOD;}return res;}

const int maxn=100+10;
const LL MOD=1e9+7;
LL dp[maxn],dp2[maxn];

int main()
{
    int i,j,n,k,d;
    while(scanf("%d%d%d",&n,&k,&d)!=EOF){
        memset(dp,0,sizeof(dp));
        memset(dp2,0,sizeof(dp2));
        dp[0]=1;
        for(i=1;i<=n;i++){
            for(j=1;j<=k;j++){
                if(i-j>=0){
                    dp[i]+=dp[i-j];
                    dp[i]%=MOD;
                }
            }
        }
        dp2[0]=1;
        for(i=1;i<=n;i++){
            for(j=1;j<d;j++){
                if(i-j>=0){
                    dp2[i]+=dp2[i-j];
                    dp2[i]%=MOD;
                }
            }
        }
        LL res=(dp[n]-dp2[n])%MOD;
        if(res<0) res+=MOD;
        printf("%I64d\n",res);
    }
    return 0;
}