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*).
n,k,d=map(int,input().split())
sto=[0]*(n+1)
sto2=[0]*(n+1)
check=[False]*(n+1)
check2=[False]*(n+1)
def dp (sum,k,sto,check):
    if(sum<0) :
        return 0
    if(sum==0):
        return 1
    if(check[sum]==True):
        return sto[sum]
    out=0
    for i in range (1,k+1):
        out=out+dp(sum-i,k,sto,check)
    sto[sum]=out
    check[sum]=True
    return out

def dplim (sum,k,d,sto2,check2):
    if(sum<0) :
        return 0
    if(sum==0):
        return 1
    if (check2[sum] == True):
        return sto2[sum]
    out=0
    for i in range (1,d):
        out=out+dplim(sum-i,k,d,sto2,check2)
    sto2[sum]=out
    check2[sum]=True
    return out
dp(n,k,sto,check)
dplim(n,k,d,sto2,check2)
x=(sto[n]-sto2[n])%1000000007
print(x)