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>

using namespace std;

int main() {
    const int Mod = 1e9 + 7;

    int n, k, d;
    cin >> n >> k >> d;

    vector<vector<int>> dp(n + 1, vector<int> (2, 0));
    dp[0][0] = 1;

    for (int i = 1 ; i <= n ; i++) {
        dp[i][0] = dp[i][1] = 0;

        for (int j = 1 ; j <= k ; j++) {
            if (i - j < 0) {
                break;
            } else {
                if (j >= d) {
                    dp[i][1] = (dp[i][1] + dp[i - j][0]) % Mod;
                    dp[i][1] = (dp[i][1] + dp[i - j][1]) % Mod;
                } else {
                    dp[i][0] = (dp[i][0] + dp[i - j][0]) % Mod;
                    dp[i][1] = (dp[i][1] +  dp[i - j][1]) % Mod;
                }
            }
        }
    }

    cout << dp[n][1];
}
# Fri Jul 07 2023 18:52:36 GMT+0300 (Moscow Standard Time)
