Problem:
Score : 400 points

Problem Statement

Given are three integers N, K, and S.

Find a sequence A_1, A_2, ..., A_N of N integers between 1 and 10^9 (inclusive) that satisfies the condition below.
We can prove that, under the conditions in Constraints, such a sequence always exists.

There are exactly K pairs (l, r) of integers such that 1 \leq l \leq r \leq N and A_l + A_{l + 1} + \cdots + A_r = S.

Constraints

1 \leq N \leq 10^5

0 \leq K \leq N

1 \leq S \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K S

Output

Print a sequence satisfying the condition, in the following format:

A_1 A_2 ... A_N

Sample Input 1

4 2 3

Sample Output 1

1 2 3 4

Two pairs (l, r) = (1, 2) and (3, 3) satisfy the condition in the statement.

Sample Input 2

5 3 100

Sample Output 2

50 50 50 30 70

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#define _GLIBCXX_DEBUG//TLEの原因になるので注意！！！！！！！！！！！
#include<cmath>
#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
vector<int> arr;
stack<int> st;
queue<int> qu;
queue<pair<int,int> > qu2;
priority_queue<int> pq;
#define rep(i,n) for(int i=0; i<(int)(n); i++)
#define rep2(i,n) for(int i=1; i<=(int)(n); i++)
#define mins(x,y) x=min(x,y)
#define maxs(x,y) x=max(x,y)
typedef set<int> set_t;
typedef set<string> set_g;
typedef complex<double> xy_t;
static const int NIL = -1;
static const int INF = 1000000007;
#define mp make_pair
#define sz(x) int(x.xise())
#define mod 1000000007
#define reps(i,s,n) for(int i = s; i < n; i++)
#define Rreps(i,n,e) for(int i = n - 1; i >= e; --i)
#define Rrep(i,n) Rreps(i,n,0)
deque<int> deq;
#define fi first
#define se second
//#define int long long

//cerrはコメントアウトしなくても通る//ただしTLEを引き起こすことがある

ll N,M,H,W,Q,A,B;
string S;
//const ll MOD = 998244353;
const ll MOD = (1e+9) + 7;
//const ll INF = 1LL << 60;
//typedef pair<ll,ll> P;
typedef pair<int, int> P;
bool chmx(int &a, int b) {
	if (b > a) {
		a = b;
		return true;
	}
	return false;
}

bool chmn(int &a, int b) {
	if (b < a) {
		a = b;
		return true;
	}
	return false;
}

int x[100005],l[100005];
bool rem[100005];

int main(){
    int n,k,s;
    cin >> n>>k>>s;
    int A[n];
    if(s%(k-1)==0){
        rep2(i,2*(k-1)){
            A[i-1]=s/(k-1);
        }
        reps(i,2*(k-1),n){
            A[i]=1000000000;
        }
    }else{
        rep(i,k){
            A[i]=s/k;
        }
        A[k]=s-k*(s/k);
        reps(i,k,2*k+1){
            A[i]=s/k;
        }
        reps(i,2*k+1,n){
            A[i]=1000000000;
        }
    }
    rep(i,n){
        cout<<A[i]<<' ';
    }
    cout<<endl;
    return 0;
}