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*).
mod = 10**9+7
def count(coins, n, sum):
    # We need sum+1 rows as the table is constructed
    # in bottom up manner using the base case 0 value
    # case (sum = 0)
    table = [[0 for x in range(n)] for x in range(sum+1)]
 
    # Fill the entries for 0 value case (n = 0)
    for i in range(n):
        table[0][i] = 1
 
    # Fill rest of the table entries in bottom up manner
    for i in range(1, sum+1):
        for j in range(n):
 
            # Count of solutions including coins[j]
            x = table[i - coins[j]][j] if i-coins[j] >= 0 else 0
 
            # Count of solutions excluding coins[j]
            y = table[i][j-1] if j >= 1 else 0
 
            # total count
            table[i][j] = x + y
 
    return table[sum][n-1]

n,k,d = map(int,input().split())
n = n-d 
coins = list(range(1,n+1))
res = count(coins,len(coins),n)%mod
print(res)
