模拟

Walking Between Houses

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There are nn houses in a row. They are numbered from 11 to nn in order from left to right. Initially you are in the house 11.

You have to perform kk moves to other house. In one move you go from your current house to some other house. You can't stay where you are (i.e., in each move the new house differs from the current house). If you go from the house xx to the house yy, the total distance you walked increases by |x−y||x−y| units of distance, where |a||a| is the absolute value of aa. It is possible to visit the same house multiple times (but you can't visit the same house in sequence).

Your goal is to walk exactly ss units of distance in total.

If it is impossible, print "NO". Otherwise print "YES" and any of the ways to do that. Remember that you should do exactly kk moves.

Input

The first line of the input contains three integers nn, kk, ss (2≤n≤1092≤n≤109, 1≤k≤2⋅1051≤k≤2⋅105, 1≤s≤10181≤s≤1018) — the number of houses, the number of moves and the total distance you want to walk.

Output

If you cannot perform kk moves with total walking distance equal to ss, print "NO".

Otherwise print "YES" on the first line and then print exactly kk integers hihi (1≤hi≤n1≤hi≤n) on the second line, where hihi is the house you visit on the ii-th move.

For each jj from 11 to k−1k−1 the following condition should be satisfied: hj≠hj+1hj≠hj+1. Also h1≠1h1≠1 should be satisfied.

Examples

input

10 2 15

output

YES
10 4 

input

10 9 45

output

YES
10 1 10 1 2 1 2 1 6 

input

10 9 81

output

YES

10 1 10 1 10 1 10 1 10

input

10 9 82

output

NO

题意:给你n个房子,你从1开始,给你k次行走,问你能不能走距离为s的路 能的话 输出 每次到达的房子.

思路:很水的一道题,结果还是没想出来,捞的嘛..... 不谈.  当k > s 时 肯定不行 就是每次都移动一步 都多了 所以不行,当 k 乘上最大路程 即n-1 小于s时肯定也不行,因为 太少了. 接下来就是判断能走到了, 可以先用 s/k 代表的是每次 走的步数, 但是 s不能整除k的时候s%k 多出来的步数 我们可以加到s/k中  就是每次 都加 1 步 知道 多出来的加完,用个 vector 然后就是判断了

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
vector < int > v;
int a[1200005];
int main()
{
    ll n,k,s;
    scanf("%lld%lld%lld",&n,&k,&s);
    if(k > s|| k*(n - 1) < s) printf("NO\n");
    else{
        cout << "YES"<<endl;
        int ans = s/k;
        int sum = s%k;
        int t;
        for(int i = 0; i < sum; i++){
            v.push_back(ans+1);
        }
        for(int i = 0; i < k - sum; i++){
            v.push_back(ans);
        }
        int p = 1;
        int q = 0;
        for(int i = 0; i < v.size(); i++){
            if(p + v[i] <= n){
               a[q++] = p + v[i];
               p += v[i];
            }
            else{
                a[q++] = p - v[i];
                p-= v[i];
            }
        }
        for(int i = 0; i < q; i++){
            printf("%d%c",a[i],i == q - 1?'\n':' ');
        }
    }

}