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: Wrong Answer)

#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <fstream>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <stack>
#include <vector>
using namespace std;
typedef long long ll;
int main(){
  int n,k;
  ll s;
  cin >> n >> k >> s;

  if(k == 0){
    if(s < 1000000000){
      for(int i = 0;i < n;i++){
        cout << 1000000000 << endl;
      }
    }
    else{
      for(int i = 0;i < n;i++){
        cout << 1 << endl;
      }
    }

    return 0;
  }
  if(k == n){
    for(int i = 0;i < n;i++)cout << s << " ";
    cout << endl;
    return 0;
  }

  vector<ll> ans(n);
  ans[0] = s - s/2;
  for(int i = 1;i <= k;i++){
    ans[i] = s - ans[i-1]; 
  }

  for(int i = k + 1;i < n;i++){
    ans[i] = s - ans[k] + 1;
  }

  for(int i = 0;i < n;i++){
    cout << ans[i] << endl;
  }
}