B - SaleBob 买电视啦！

 

B -Sale _p

Time Limit: 2000MS Memory Limit: 262144K 64bit IO Format: %I64d& %I64u

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Description

Once Bob got to a sale of old TV sets. Therewere n TV sets at thatsale. TV set with index i costs a_i bellars. SomeTV sets have a negative price — their owners are ready to pay Bobif he buys their useless apparatus. Bob can «buy» any TV sets hewants. Though he's very strong, Bob can carry atmost m TV sets, and hehas no desire to go to the sale for the second time. Please, helpBob find out the maximum sum of money that he can earn.

Input

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

Output

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

Sample Input

Input

5 3-6 0 35 -2 4

Output

8

Input

4 27 0 0 -7

Output

7

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#include<iostream>#include<algorithm>#include<cmath>using namespace std;int main(){        int n,m,a[100],i,sun;        while(scanf("%d %d",&n,&m)!=EOF)        {                sun=0;                for(i=0;i<n;i++)                        cin>>a[i];                sort(a,a+n);                for(i=0;i<m;i++)                {                        if(a[i]<0)                                sun+=a[i];                 }                cout<<abs(sun)<<endl;        }        return 0;}