Codeforces 34 B. Sale

题目链接：http://codeforces.com/contest/34/problem/B

Once Bob got to a sale of old TV sets. There were n TV sets at that sale. TV set with index i costs ai bellars. Some TV sets have a negative price — their owners are ready to pay Bob if he buys their useless apparatus. Bob can «buy» any TV sets he wants. Though he's very strong, Bob can carry at most m TV sets, and he has no desire to go to the sale for the second time. Please, help Bob find out the maximum sum of money that he can earn.

Input

The first line contains two space-separated integers n and m (1 ≤ m ≤ n ≤ 100) — amount of TV sets at the sale, and amount of TV sets that Bob can carry. The following line contains n space-separated integers ai ( - 1000 ≤ ai ≤ 1000) — prices of the TV sets.

Output

Output the only number — the maximum sum of money that Bob can earn, given that he can carry at most m TV sets.

Examples

input

Copy

5 3
-6 0 35 -2 4

output

Copy

8

input

Copy

4 2
7 0 0 -7

output

Copy

7

题目大意：

给定n台电视机价格（有负数，为什么？我也不知道为什么有负数），Bob最多可以买m台，问Bob可以获得的最大价值。

题目思路：

把价格取反，排序之后把所有正数加起来即可，注意，最多m个。

代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<queue>
#include<stack>
#include<map>

using namespace std;

#define FOU(i,x,y) for(int i=x;i<=y;i++)
#define FOD(i,x,y) for(int i=x;i>=y;i--)
#define MEM(a,val) memset(a,val,sizeof(a))
#define PI acos(-1.0)

const double EXP = 1e-9;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const ll MINF = 0x3f3f3f3f3f3f3f3f;
const double DINF = 0xffffffffffff;
const int mod = 1e9+7;
const int N = 1e6+5;

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    std::ios::sync_with_stdio(false);
    int n,m;
    int a[105];
    cin>>n>>m;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
        a[i] = -a[i];
    }
    sort(a+1,a+1+n);
    int ans=0;
    for(int i=n;i>=1&&i>n-m;i--)
    {
        if(a[i]>0)
            ans+=a[i];
        else
            break;
    }
    cout<<ans<<endl;

    return 0;
}