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

4
2 4 4 2

output

8

input

4
2 2 3 5

output

0

input

4
100003 100004 100005 100006

output

10000800015

题意：有N条木棍，求用这n条木棍可得到的所有矩形的最大面积（其中木棍的长度可以减少1）

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define LL long long
using namespace std;
int a[100000+10];
int main()
{
int n,i,j,k,l;
while(scanf("%d",&n)!=EOF)
{
memset(a,0,sizeof(a));
for(i=1;i<=n;i++)
scanf("%d",&a[i]);
sort(a+1,a+n+1);
bool flag=false;
LL ans=0;
for(i=n-1;i>=1;i--)
{
if(a[i+1]-a[i]<2)
{
if(!flag)
{
k=a[i];
flag=true;
}
else
{
ans+=(LL)k*a[i];
flag=false;
}
i--;
}
}
printf("%lld\n",ans);
}
return 0;
}