Codeforces 431C k-Tree (基础dp)

 k-Tree

time limit per test：1 second

memory limit per test：256 megabytes

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 ak-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 (exactlyk 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 weightn (the sum of all weights of the edges in the path) are there, starting from the root of ak-tree and also containing at least one edge of weight at leastd?".

Help Dima find an answer to his question. As the number of ways can be rather large, print it modulo1000000007 (10⁹ + 7).

Input

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

Output

Print a single integer — the answer to the problem modulo1000000007 (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

题目链接：http://codeforces.com/problemset/problem/431/C

题目大意：一棵k-Tree表示一棵满k叉树，每个祖先到孩子的边权依次为1-k，现在要求一条和为n的路径，要求路径上至少有一条边的权值大于等于d，求满足条件的路径个数

题目分析：类比背包，容易想到状态dp[i][j]表示合到数字i时，是否有一条边权大于等于d的路径数，转移不说了

#include <cstdio>int const MOD = 1e9 + 7;int dp[105][2];int main(){    int n, k, d;    scanf("%d %d %d", &n, &k, &d);    dp[0][0] = 1;    for(int i = 0; i < n; i++)    {        for(int j = 1; j <= k && i + j <= n; j++)        {            if(j < d)                dp[i + j][0] = (dp[i + j][0] % MOD + dp[i][0] % MOD) % MOD;            else                dp[i + j][1] = (dp[i + j][1] % MOD + dp[i][0] % MOD) % MOD;            dp[i + j][1] = (dp[i + j][1] % MOD + dp[i][1] % MOD) % MOD;        }    }    printf("%d\n", dp[n][1]);}