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 = int(1e9 + 7)


def solve():
    n, k, d = map(int, input().split())

    dp = []
    dp2 = []
    for _ in range(n + 1):
        dp.append([0 for __ in range(n + 1)])
        dp2.append([0 for __ in range(n + 1)])

    dp[0][0] = 1
    ans = 0
    for i in range(1, n + 1):
        for j in range(1, n + 1):
            for p in range(1, k + 1):
                if j - p >= 0:
                    dp[i][j] += dp[i - 1][j - p]
                    dp[i][j] %= MOD

        ans += dp[i][n]
        ans %= MOD

    dp2[0][0] = 1
    ans2 = 0
    for i in range(1, n + 1):
        for j in range(1, n + 1):
            for p in range(1, d):
                if j - p >= 0:
                    dp2[i][j] += dp2[i - 1][j - p]
                    dp2[i][j] %= MOD

        ans2 += dp2[i][n]
        ans2 %= MOD

    print((ans + MOD - ans2) % MOD)


if __name__ == "__main__":
    solve()
