#include<cstdio>

#include<cstring>

#include<iostream>

#include<cmath>

#include<algorithm>

using namespace std;

#define rep(i,n) for(int i=1;i<=n;i++)

#define clr(x,c) memset(x,c,sizeof(x))

#define REP(i,s,t) for(int i=s;i<=t;i++)

#define op() clr(head,0);pt=edges;

#define qwq(x) for(edge *o=head[x];o;o=o->next)

int read(){

	int x=0;char c=getchar();bool f=true;

	while(!isdigit(c)) {

		if(c=='-') f=false;c=getchar();

	}

	while(isdigit(c)) x=x*10+c-'0',c=getchar();

	return x;

}

const int nmax=3005;

const int maxn=6000005;

const int inf=0x7f7f7f7f;

struct edge{

	int to,cap;edge *next,*rev;

};

edge edges[maxn],*pt,*head[nmax],*p[nmax],*cur[nmax];

int cnt[nmax],h[nmax],g[nmax];

void add(int u,int v,int w){

	pt->to=v;pt->cap=w;pt->next=head[u];head[u]=pt++;

}

void adde(int u,int v,int w){

	add(u,v,w);add(v,u,0);head[u]->rev=head[v];head[v]->rev=head[u];

}

int maxflow(int s,int t,int n){

	clr(cnt,0);cnt[0]=n;clr(h,0);

	int flow=0,a=inf,x=s;edge *e;

	while(h[s]<n){

		for(e=cur[x];e;e=e->next) if(e->cap>0&&h[x]==h[e->to]+1) break;

		if(e){

			a=min(a,e->cap);p[e->to]=cur[x]=e;x=e->to;

			if(x==t){

				while(x!=s) p[x]->cap-=a,p[x]->rev->cap+=a,x=p[x]->rev->to;

				flow+=a,a=inf;

			}

		}else{

			if(!--cnt[h[x]]) break;

			h[x]=n;

			for(e=head[x];e;e=e->next) if(e->cap>0&&h[x]>h[e->to]+1)

			  h[x]=h[e->to]+1,cur[x]=e;

			cnt[h[x]]++;

			if(x!=s) x=p[x]->rev->to;

		}

	}

	return flow;

}

int gcd(int x,int y){

	return y==0?x:gcd(y,x%y);

}

bool check(int x,int y){

	if(gcd(x,y)!=1) return false;

	int tmp=x*x+y*y,temp=sqrt(tmp);

	if(temp*temp==tmp) return true;

	return false;

}

int main(){

	op();

	int n=read(),s=0,t=n+1,ans=0;

	rep(i,n){

		g[i]=read();ans+=g[i];

		g[i]%2?adde(s,i,g[i]):adde(i,t,g[i]);

	}

	rep(i,n) if(g[i]%2) rep(j,n) if(!(g[j]%2))

	   if(check(g[i],g[j])) adde(i,j,inf);

	/*REP(i,s,t) {

		qwq(i) printf("%d ",o->to);printf("\n");

	}*/

	printf("%d\n",ans-maxflow(s,t,t+1));

	return 0;

}

有N个正整数，需要从中选出一些数，使这些数的和最大。
若两个数a,b同时满足以下条件，则a,b不能同时被选
1:存在正整数C，使a*a+b*b=c*c
2:gcd(a,b)=1

n<=3000。

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