#include <bits/stdc++.h>
using namespace std;
typedef signed long long ll;

#undef _P
#define _P(...) (void)printf(__VA_ARGS__)
#define FOR(x,to) for(x=0;x<to;x++)
#define ITR(x,c) for(__typeof(c.begin()) x=c.begin();x!=c.end();x++)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
//-------------------------------------------------------

int N,D,K;

int ok[101][1001][11];
vector<int> memo[101][1001][11];

int dfs(int n,int d,int k) {
	if(d==0 && k==0) return 1;
	if(n>N) return 0;
	if(k<=0) return 0;
	if(d<n) return 0;
	if(ok[n][d][k]>=0) return ok[n][d][k];
	ok[n][d][k]=0;
	
	if(dfs(n+1,d-n,k-1)==1) {
		ok[n][d][k]=1;
		memo[n][d][k].push_back(n);
		ITR(it,memo[n+1][d-n][k-1]) memo[n][d][k].push_back(*it);
	}
	else if(dfs(n+1,d,k)==1) {
		ok[n][d][k]=1;
		ITR(it,memo[n+1][d][k]) memo[n][d][k].push_back(*it);
	}
	return ok[n][d][k];
}


void solve() {
	int i,j,k,l,r,x,y; string s;
	cin>>N>>D>>K;
	
	MINUS(ok);
	x=dfs(1,D,K);
	if(x==0) _P("-1\n");
	else {
		FOR(i,memo[1][D][K].size()) {
			if(i) _P(" ");
			_P("%d",memo[1][D][K][i]);
		}
		_P("\n");
	}
}


int main(int argc,char** argv){
	string s;int i;
	if(argc==1) ios::sync_with_stdio(false);
	FOR(i,argc-1) s+=argv[i+1],s+='\n';
	FOR(i,s.size()) ungetc(s[s.size()-1-i],stdin);
	solve(); return 0;
}