Ehab and subtraction(思维题)

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You're given an array aa. You should repeat the following operation kk times: find the minimum non-zero element in the array, print it, and then subtract it from all the non-zero elements of the array. If all the elements are 0s, just print 0.

Input

The first line contains integers nn and kk (1≤n,k≤105)(1≤n,k≤105), the length of the array and the number of operations you should perform.

The second line contains nn space-separated integers a1,a2,…,ana1,a2,…,an (1≤ai≤109)(1≤ai≤109), the elements of the array.

Output

Print the minimum non-zero element before each operation in a new line.

Examples

input

Copy

3 5
1 2 3

output

Copy

1
1
1
0
0

input

Copy

4 2
10 3 5 3

output

Copy

3
2

Note

In the first sample:

In the first step: the array is [1,2,3][1,2,3], so the minimum non-zero element is 1.

In the second step: the array is [0,1,2][0,1,2], so the minimum non-zero element is 1.

In the third step: the array is [0,0,1][0,0,1], so the minimum non-zero element is 1.

In the fourth and fifth step: the array is [0,0,0][0,0,0], so we printed 0.

In the second sample:

In the first step: the array is [10,3,5,3][10,3,5,3], so the minimum non-zero element is 3.

In the second step: the array is [7,0,2,0][7,0,2,0], so the minimum non-zero element is 2.

题解：如果直接暴力去找的话必然是会超时的，我们可以利用c++排序的功能排序好，然后进行比较每次输出最小的就大大提高了效率，从而不会是超时

代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>

using namespace std;

int main()
{
	
	int n,m;
	cin>>n>>m;
	int a[100005];
	for(int t=0;t<n;t++)
	{
		scanf("%d",&a[t]);
	}
    sort(a,a+n);
    int k=0;
    int temp=0;
    for(int t=0;t<m;t++)
    {
    	while(k!=n-1&&a[k]==temp)k++;
    	printf("%d\n",a[k]-temp);
    	temp=a[k];
	}
	return 0;
}