Like everyone, cows enjoy variety. Their current fancy is new shapes for pastures. The old rectangular shapes are out of favor; new geometries are the favorite.

I. M. Hei, the lead cow pasture architect, is in charge of creating a triangular pasture surrounded by nice white fence rails. She is supplied with N (3 <= N <= 40) fence segments (each of integer length Li (1 <= Li <= 40) and must arrange them into a triangular pasture with the largest grazing area. Ms. Hei must use all the rails to create three sides of non-zero length.

Help Ms. Hei convince the rest of the herd that plenty of grazing land will be available. Calculate the largest area that may be enclosed with a supplied set of fence segments.

* Line 1: A single integer N
* Lines 2..N+1: N lines, each with a single integer representing one fence segment's length.  The lengths are not necessarily unique.

A single line with the integer that is the truncated integer representation of the largest possible enclosed area multiplied by 100. Output -1 if no triangle of positive area may be constructed.

#include<iostream>
#include<math.h>
#include<string.h>
#include<algorithm>
#include<stdio.h>
using namespace std;
bool flag[810][810];int main() {int n,i,j,k,ans;int str[110];while(~scanf("%d",&n)) {ans = 0;for(i=1; i<=n; i++) {scanf("%d",&str[i]);ans+=str[i];}memset(flag,false,sizeof(flag));flag[0][0]=true;int mid=ans/2;
　　　　　//判断是否可以形成三角形for(i=1; i<=n; i++)for(j=mid; j>=0; j--)for(k=j; k>=0; k--)if(j>=str[i]&&flag[j-str[i]][k] || k>=str[i]&&flag[j][k-str[i]])flag[j][k]=true;int Max=-1;for(i=mid; i>=1; i--)for(j=i; j>=1; j--) {if(flag[i][j]) {k=ans-i-j;if(i+j>k && i+k>j && j+k>i) {double p=(i+j+k)/2.0;int c=(int)(sqrt(p*(p-i)*(p-j)*(p-k))*100);if(c>Max)Max=c;}}}printf("%d\n",Max);}return 0;
}