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*).
from sys import stdin, stdout 

n, k, d = map(int, stdin.readline().split())
# dp[sum][depth][has d?]
MOD = int(1e9 + 7)

dp = [[[0] * 2 for _ in range(n + 1)] for _ in range(n + 1)]
dp[0][0][0] = 1

for depth in range(1, n + 1):
    for sum in range(1, n + 1):
        for add in range(1, k + 1):
            if sum - add >= 0:
                dp[sum][depth][0] += dp[sum - add][depth - 1][0]
                dp[sum][depth][0] %= MOD
                if add >= d: 
                    dp[sum][depth][1] += dp[sum - add][depth - 1][0]
                    dp[sum][depth][1] %= MOD
                else:
                    dp[sum][depth][1] += dp[sum - add][depth - 1][1]
                    dp[sum][depth][1] %= MOD 

# print(dp)
ans = 0
for i in range(0, n + 1):
    ans += dp[n][i][1]
    ans %= MOD 
print(ans)
