Codeforces 525C. Ilya and Sticks

C. Ilya and Sticks

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

In the evening, after the contest Ilya was bored, and he really felt like maximizing. He remembered that he had a set of n sticks and an instrument. Each stick is characterized by its length li.

Ilya decided to make a rectangle from the sticks. And due to his whim, he decided to make rectangles in such a way that maximizes their total area. Each stick is used in making at most one rectangle, it is possible that some of sticks remain unused. Bending sticks is not allowed.

Sticks with lengths a1, a2, a3 and a4 can make a rectangle if the following properties are observed:

• a1 ≤ a2 ≤ a3 ≤ a4
• a1 = a2
• a3 = a4

A rectangle can be made of sticks with lengths of, for example, 3 3 3 3 or 2 2 4 4. A rectangle cannot be made of, for example, sticks5 5 5 7.

Ilya also has an instrument which can reduce the length of the sticks. The sticks are made of a special material, so the length of each stick can be reduced by at most one. For example, a stick with length 5 can either stay at this length or be transformed into a stick of length 4.

You have to answer the question — what maximum total area of the rectangles can Ilya get with a file if makes rectangles from the available sticks?

Input

The first line of the input contains a positive integer n (1 ≤ n ≤ 105) — the number of the available sticks.

The second line of the input contains n positive integers li (2 ≤ li ≤ 106) — the lengths of the sticks.

Output

The first line of the output must contain a single non-negative integer — the maximum total area of the rectangles that Ilya can make from the available sticks.

Examples

input

42 4 4 2

output

8

input

42 2 3 5

output

0

input

4100003 100004 100005 100006

output

10000800015

恩，题意是说，相当于是给出一系列长度的木棍，分别用4个木棍组成长方形，这里存在一种操作即将木棍的长度减1  问最终可组成的长方体的面积和。开始以为是求可组成的面积最大的长方体的面积。。。英语是硬伤

#include <iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#define maxn 100010using namespace std;int r[5];int num[maxn];int main(){    int n;    __int64 ans;    while(~scanf("%d",&n))    {        ans=0;        for(int i=0;i<n;++i)            scanf("%d",&num[i]);        sort(num,num+n);        int k=0;        memset(r,0,sizeof(r));        for(int i=n-1;i>=0;--i)        {            if(num[i]-num[i-1]<=1)            {                r[k]=r[k+1]=num[i-1];                k+=2;                i--;            }            if(k>=4)            {                ans+=(__int64)r[0]*r[2];                k=0;            }        }        printf("%I64d\n",ans);    }    return 0;}