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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define all(x)   (x).begin(),(x).end()
#define YES() printf("YES\n")
#define NO() printf("NO\n")
#define isYES(x) printf("%s\n",(x) ? "YES" : "NO")
#define Yes() printf("Yes\n")
#define No() printf("No\n")
#define isYes(x) printf("%s\n",(x) ? "Yes" : "No")
#define isIn(x,y,h,w) (x >= 0 && x < h && y >= 0 && y < w)

#define int long long
//using ll = long long;
using P = pair<int,int>;

ostream &operator<<(ostream &os,const P &p){ return os << "(" << p.first << "," << p.second << ")"; }

template<class T> T &chmin(T &a,const T &b){ return a = min(a,b); }
template<class T> T &chmax(T &a,const T &b){ return a = max(a,b); }
 
const int INF=1e+18;
const double EPS=1e-9;
const int MOD=1000000007;

const int dx[]={1,0,-1,0},dy[]={0,-1,0,1};

void output(vector<int> ans){
	for(int i = 0;i < ans.size();i++) cout << (i ? " " : "") << ans[i];
	cout << endl;
}

signed main(){
	int n,k,s;
	cin >> n >> k >> s;
	vector<int> ans(n);
	if(k == n){
		for(int i = 0;i < n;i++) ans[i] = s;
		output(ans);
		return 0;
	}
	int d = n - k;
	for(int i = 0;i < n;i++){
		if(s - d <= 0){
			ans[i] = 1000000000;
			d--;
		}
		else if(i % (d + 1) == 0) ans[i] = s - d;
		else ans[i] = 1;
	}
	output(ans);
}