D. Array Splitting(后缀数组）

You are given an array 𝑎1,𝑎2,…,𝑎𝑛 and an integer 𝑘

.

You are asked to divide this array into 𝑘

non-empty consecutive subarrays. Every element in the array should be included in exactly one subarray. Let 𝑓(𝑖) be the index of subarray the 𝑖-th element belongs to. Subarrays are numbered from left to right and from 1 to 𝑘

.

Let the cost of division be equal to ∑𝑖=1𝑛(𝑎𝑖⋅𝑓(𝑖))

. For example, if 𝑎=[1,−2,−3,4,−5,6,−7] and we divide it into 3 subbarays in the following way: [1,−2,−3],[4,−5],[6,−7], then the cost of division is equal to 1⋅1−2⋅1−3⋅1+4⋅2−5⋅2+6⋅3−7⋅3=−9

.

Calculate the maximum cost you can obtain by dividing the array 𝑎

into 𝑘

non-empty consecutive subarrays.

Input

The first line contains two integers 𝑛

and 𝑘 (1≤𝑘≤𝑛≤3⋅105

).

The second line contains 𝑛

integers 𝑎1,𝑎2,…,𝑎𝑛 (|𝑎𝑖|≤106

).

Output

Print the maximum cost you can obtain by dividing the array 𝑎

into 𝑘

nonempty consecutive subarrays.

Examples

Input

Copy

5 2
-1 -2 5 -4 8

Output

Copy

15

Input

Copy

7 6
-3 0 -1 -2 -2 -4 -1

Output

Copy

-45

Input

Copy

4 1
3 -1 6 0

Output

Copy

8

题解：将n个数的数组分成k个连续的子数组，并且第i个子数组的权值为i，则我们可以用后缀和。首先我们一定每个数都得至少取一次，则我么一定要取a[1],然后将后面的n-1个数排序，
因为题目要求获得答案最大，并且这后n-1个数我们任意取k-1个都能保证符合题意中的分法，则为了保证答案最大我们就要取比较大的前k-1个了～～

#include<bits/stdc++.h>
#include<iostream>
#include<stdio.h>
#include<iomanip>
#include<stack>
#include<queue>
#include<algorithm>
#include<cstring>
#include<map>
#include<vector>
#include<numeric>
#include<iterator>
#include<cmath>
#define mem(a,x) memset(a,x,sizeof(a));
using namespace std;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a%b); }
int lcm(int a, int b) { return a * b / gcd(a, b); }
const int INF = 0x3f3f3f3f;
typedef long long ll;
const int mod=1000000007;
typedef pair<int,int>Pi;
typedef pair<ll, ll>Pii;
map<int,int>mp;
map<int, char *>mp1;
map<char *, int>mp2;
map<char, int>mp3;
map<string,int>mp4;
map<char,int>mp5;
const int maxn = 300010;
ll a[maxn];

int read(){
    int flag=1;
    int sum=0;
    char c=getchar();
    while(c<'0'||c>'9'){
        if(c=='-')flag=-1;
        c=getchar();
    }
    while(c>='0'&&c<='9'){
        sum=sum*10+c-'0';
        c=getchar();
    }
    return sum*flag;
}
ll Read(){
    int flag=1;
    ll sum=0;
    char c=getchar();
    while(c<'0'||c>'9'){
        if(c=='-')flag=-1;
        c=getchar();
    }
    while(c>='0'&&c<='9'){
        sum=sum*10+c-'0';
        c=getchar();
    }
    return sum*flag;
}
ll quickmul(ll a,ll b){
    ll ans=0;
    while(b){
        if(b&1){
            ans=(ans+a)%mod;
        }
        a=(a+a)%mod;
        b>>=1;
    }
    return ans;
}
ll quickpow(ll a,ll b){
    ll ans=1;
    while(b){
        if(b&1)
            ans=(ans*a)%mod;
        a=(a*a)%mod;
        b>>=1;
    }
    return ans;
}
int main()
{
    int n,k;
    cin>>n>>k;
    for(int i=1;i<=n;i++)cin>>a[i];
    for(int i=n;i>=1;i--)a[i]+=a[i+1];
    sort(a+2,a+1+n,greater<ll>());
    ll ans=0;
    for(int i=1;i<=k;i++)ans+=a[i];
    cout<<ans<<endl;
    return 0;
}