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*).
def ways(sum1,d,k,dp):
    if sum1==0:
        return 1
    if dp[0][sum1]!=0:
        return dp[0][sum1]
    for i in range(1,k+1):
        if sum1-i<0:
            break
        a=ways(sum1-i,d,k,dp)
        dp[0][sum1]+=a
        
    return dp[0][sum1]

def ways1(sum1,d,k,dp,count):
    if sum1==0 and count>0:
        return 1
    if sum1==0 and count==0:
        return 0
    if dp[1][sum1]!=0 and count==0:
        return dp[1][sum1]
    if dp[1][sum1]!=0 and count>0:
        return dp[0][sum1]
     
    for i in range(1,k+1):
        if sum1-i<0:
            break
        if i>=d:
            count+=1
        a=ways1(sum1-i,d,k,dp,count)
        if i>=d:
            count-=1
        dp[1][sum1]+=a
        
    return dp[1][sum1]

sum1,k,d=[int(i) for i in input().split()]
dp=[]
for i in range(2):
    dp.append([0]*(sum1+1))
ways(sum1,d,k,dp)
ways1(sum1,d,k,dp,0)
print(dp[1][-1])
