D - MaratonIME in the golden moment Gym - 101375D

本文介绍了一种计算方法,用于确定当所有成员参加时,船的总能力。该方法通过计算每两个成员间的能力乘积之和来评估团队的整体表现。

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D - MaratonIME in the golden moment

 Gym - 101375D 

It is a common knowledge the importance in practicing a physical activity, mainly when you take part in ICPC competitions. Keeping this in mind, Giovana Delfino invited her friends from MaratonIME to go rowing.

The initial total ability of the boat is 0. Despite the equal enthusiasm of all members, each friend i has a physical ability h(i). Dear teacher Gabi knows that whenever two friends x and y are together in the boat, the boat's total ability is increased by h(xh(y).

Giovana invited n friends, but rowing is not always possible when you have programming assignments and exams of the great professor Arnaldo Mandel ahead, so sometimes not everyone shows up. However, the teacher Gabi is very optimistic and hopes that, for the final semester competition, all n friends will attend in order to increase MaratonIME odds. Help Gabi find the total ability of the boat in this possible golden moment, assuming all friends show up.

Input

The first line contains one integer n (1 ≤ n ≤ 105) - the number of friends invited to row.

The second line contains n integers h1, h2, ..., hn (1 ≤ hi ≤ 3·104) - the abilities of each member.

Output

Print one integer - the boat's total ability in the golden moment.

Examples

Input

2
5 10

Output

50

Input

4
1 3 7 11

Output

152

Note

In the first example, in the golden moment there will be two members on the boat, then the ability there will be 5 × 10 = 50.


#include <stdio.h>
#include <stdlib.h>

int main()
{
    long long int n,s,a[100001],b[100001];
    int i;
    scanf("%lld",&n);
    s=0;
    for(i=0;i<n;i++)
    {
        scanf("%lld",&a[i]);
        if(i==0) b[i]=a[i];
        else b[i]=b[i-1]+a[i];
    }
    if(n==1)s=0;
    else
    {
        for(i=0;i<n;i++)
        {
            s=s+(b[i]-a[i])*a[i];
        }
    }
    printf("%lld\n",s);
    return 0;
}

 

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