Petya and Inequiations 题解

Little Petya loves inequations. Help him find n positive integers a₁, a₂, ..., a_n, such that the following two conditions are satisfied:

• a₁² + a₂² + ... + a_n² ≥ x
• a₁ + a₂ + ... + a_n ≤ y

Input

The first line contains three space-separated integers n, x and y (1 ≤ n ≤ 10⁵, 1 ≤ x ≤ 10¹², 1 ≤ y ≤ 10⁶).

Please do not use the %lld specificator to read or write 64-bit integers in С++. It is recommended to use cin, cout streams or the %I64d specificator.

Output

Print n positive integers that satisfy the conditions, one integer per line. If such numbers do not exist, print a single number "-1". If there are several solutions, print any of them.

Example

Input

5 15 15

Output

44112

Input

2 3 2

Output

-1

Input

1 99 11

Output

11

思路：

令a₁ + a₂ + ... + a_n= y，那么现在想办法如何拆分y使得a₁² + a₂² + ... + a_n²取得最大值，当其中n-1个数都取1，剩下一个数取y-（n-1）时a₁² + a₂² + ... + a_n²取得最大值。因此，令a₁ + a₂ + ... + a_n= y时，若a₁² + a₂² + ... + a_n²大于等于x，那么存在解，且此时的a1,…,an是其中的一组解，否则不存在解。

代码：

#include <iostream>using namespace std;int main(){    int64_t n,y;    int64_t x;    while(cin>>n>>x>>y)    {        if(y>=n)        {            int64_t temp=y-(n-1);            if(temp*temp+(n-1)>=x)            {                cout<<temp<<endl;                for(int i=0;i<n-1;i++)                    cout<<1<<endl;            }            else                cout<<-1<<endl;        }        else            cout<<-1<<endl;    }    return 0;}