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*).
import sys
input = sys.stdin.readline
num = 1000000007

n, k, d = [int(x) for x in input().split()]
dp = [0]*(n+1)
 
for i in range(1, n+1):
    if i <= k:
        dp[i] = 1
    for j in range(1, k+1):
        if (i-j) > 0:
            dp[i] += dp[i-j]
            
dp1 = [0]*(n+1)

# print(dp1)
    
for i in range(1, n+1):
    if ((i <= k) and (i < d)):
        dp1[i] = 1
    for j in range(1, k+1):
        if j >= d:
            continue
        if (i-j) > 0:
            dp1[i] += dp1[i-j]
        # print(i, j, '>>', dp1)
            
# print(dp1)
# print(dp)
print((dp[-1] - dp1[-1])%num)
        
    
