codeforces 231B Magic, Wizardry and Wonders（超强技巧+思维+观察力）

Vasya the Great Magician and Conjurer loves all kinds of miracles and wizardry. In one wave of a magic wand he can turn an object into something else. But, as you all know, there is no better magic in the Universe than the magic of numbers. That's why Vasya adores math and spends a lot of time turning some numbers into some other ones.

This morning he has n cards with integers lined up in front of him. Each integer is not less than 1, but not greater than l. When Vasya waves his magic wand, two rightmost cards vanish from the line and a new card magically appears in their place. It contains the difference between the left and the right numbers on the two vanished cards. Vasya was very interested to know what would happen next, and so he waved with his magic wand on and on, until the table had a single card left.

Suppose that Vasya originally had the following cards: 4, 1, 1, 3 (listed from left to right). Then after the first wave the line would be: 4, 1, -2, and after the second one: 4, 3, and after the third one the table would have a single card with number 1.

Please note that in spite of the fact that initially all the numbers on the cards were not less than 1 and not greater than l, the numbers on the appearing cards can be anything, no restrictions are imposed on them.

It is now evening. Vasya is very tired and wants to return everything back, but does not remember which cards he had in the morning. He only remembers that there were ncards, they contained integers from 1 to l, and after all magical actions he was left with a single card containing number d.

Help Vasya to recover the initial set of cards with numbers.

Input

The single line contains three space-separated integers: n (2 ≤ n ≤ 100) — the initial number of cards on the table, d (|d| ≤ 10⁴) — the number on the card that was left on the table after all the magical actions, and l (1 ≤ l ≤ 100) — the limits for the initial integers.

Output

If Vasya is mistaken, that is, if there doesn't exist a set that meets the requirements given in the statement, then print a single number -1, otherwise print the sought set containing n integers from 1 to l. Separate the integers by spaces. Print the integers in the order, in which they were written on the cards from left to right. If there are several suitable sets of numbers, you can print any of them.

Example

Input

3 3 2

Output

2 1 2 

Input

5 -4 3

Output

-1

Input

5 -4 4

Output

2 4 1 4 1 

 【题解】 刚开始看着这题很懵，有点不知所措，然后第一反应是用搜索，不过看到100的数据就放弃了，，其实有可能是可以的，看数据水不水，因为只要找到一组满足题意的就直接退出，所以2s.....不知道可不可以，，有兴趣试一下。

 就是说给一行数，从最后面的两个开始，倒数第二个减去倒数第一个的值放回到队尾，重复这种操作直到队中只剩一个元素时，输出其值。

 这一行数的范围是1~l；

仔细！！！观察可以发现，所有奇数和 减去 偶数和 等于最后的值d，不过这没用，重点是给定的最后的值d是在最大值与最小值之间的，最大值就是奇数项全取最大值，偶数项全取最小值后的差，最小值是奇数项全取最小值而偶数项全取最大值的差（前面减后面），如果d在这个范围内，那一定能找到，否则直接输出-1，判断了一定能找到后，就是枚举的过程，最大的减一减，最小的加一加，寻找满足条件的数，具体过程看代码：

【AC代码】

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std;const int N=105;int main(){    int n,d,L,a[N],b[N],m1,m2;    while(~scanf("%d%d%d",&n,&d,&L))    {        if(n%2)//奇偶项的个数        {            m1=n/2+1;            m2=n/2;        }        else m1=m2=n/2;        for(int i=0;i<m1;++i) //奇数项全取最大值            a[i]=L;        for(int i=0;i<m2;++i)//偶数项全取最小值            b[i]=1;        int d1=L*m1-1*m2;//求范围        int d2=1*m1-L*m2;        if(d>d1 || d<d2 )//不在范围        {            printf("-1\n");            return 0;        }        for(int i=0;i<m1;i++)//初始化全部项，准备加加减减            a[i]=L;        for(int i=0;i<m2;i++)            b[i]=1;        int p=0;        d1=L*m1;        d2=1*m2;        while(1)        {            if(d1-d2 == d) break;            while(a[p]==1) p++;            if(p==m1) break;            a[p]--;            d1--;        }        if(p==m1)        {            p=0;            while(1)            {                if(d1-d2 == d ) break;                while(b[p]==L) p++;                if(p==m2) break;                b[p]++;                d2++;            }        }        int i;        for(i=0;i<m2;++i)            printf("%d %d ",a[i],b[i]);//奇偶相间输出        if(m1>m2)            printf("%d",a[i]);        printf("\n");    }}