// SEQCOUNT Solution. Written by Sergey Kulik

#include <cstdio>
#include <algorithm>
using namespace std;

#define mod 1000000007
#define maxn 100000 + 5

int n, m, k, t, i, j, dp[2][maxn], ac[maxn], l;

void upd (int &a, int b) {
	a += b;
	if (a >= mod) a -= mod;
}

int main () {
	scanf("%d %d %d", &n, &m, &k);
	for(i = 1; i <= m; i++) t += i;
	if (t > n) { // If the minimal possible sum is greater than N, there is no way to construct a nice sequence
		puts("0");
		return 0;
	}
	for(i = 1; i <= n; i++) dp[0][i] = 0;
	dp[0][0] = 1;
	for(i = 1; i <= m; i++) {
		for(j = 0; j < m - i + 1; j++) ac[j] = 0;
		for(j = 0; j <= n; j++) dp[1 - l][j] = 0;
		for(j = 0; j <= n; j++) {
			dp[1 - l][j] = ac[j % (m - i + 1)]; // all the requred summands for this state will have equal remainder modulo (m - i + 1)
			upd(ac[j % (m - i + 1)], dp[l][j]); // add the current state for the further updates
			if (j - (m - i + 1) * (k) >= 0) upd(ac[j % (m - i + 1)], mod - dp[l][j - (m - i + 1) * (k)]); // we shouldn't consider the differences larger than k
		}
		l ^= 1;
	}
	printf("%d\n", dp[l][n]);
	return 0;
}