codeforces 743 D. Chloe and pleasant prizes(DFS)

D. Chloe and pleasant prizes

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Generous sponsors of the olympiad in which Chloe and Vladik took part allowed all the participants to choose a prize for them on their own. Christmas is coming, so sponsors decided to decorate the Christmas tree with their prizes.

They took n prizes for the contestants and wrote on each of them a unique id (integer from 1 to n).
A gift i is characterized by integer ai —
pleasantness of the gift. The pleasantness of the gift can be positive, negative or zero. Sponsors placed the gift 1 on the top of the tree.
All the other gifts hung on a rope tied to some other gift so that each gift hung on the first gift, possibly with a sequence of ropes and another gifts. Formally, the gifts formed a rooted tree with n vertices.

The prize-giving procedure goes in the following way: the participants come to the tree one after another, choose any of the remaining gifts and cut the rope this prize hang on. Note that all the ropes which were used to hang other prizes on the chosen one
are not cut. So the contestant gets the chosen gift as well as the all the gifts that hang on it, possibly with a sequence of ropes and another gifts.

Our friends, Chloe and Vladik, shared the first place on the olympiad and they will choose prizes at the same time! To keep themselves from fighting, they decided to choose two different gifts so that the sets of the gifts that hang on them with a sequence
of ropes and another gifts don't intersect. In other words, there shouldn't be any gift that hang both on the gift chosen by Chloe and on the gift chosen by Vladik. From all of the possible variants they will choose such pair of prizes that the sum of pleasantness
of all the gifts that they will take after cutting the ropes is as large as possible.

Print the maximum sum of pleasantness that Vladik and Chloe can get. If it is impossible for them to choose the gifts without fighting, print Impossible.

Input

The first line contains a single integer n (1 ≤ n ≤ 2·105) —
the number of gifts.

The next line contains n integers a1, a2, ..., an ( - 109 ≤ ai ≤ 109) —
the pleasantness of the gifts.

The next (n - 1) lines contain two numbers each. The i-th
of these lines contains integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) —
the description of the tree's edges. It means that gifts with numbers ui and vi are
connected to each other with a rope. The gifts' ids in the description of the ropes can be given in arbirtary order: vi hangs
on ui or ui hangs
on vi.

It is guaranteed that all the gifts hang on the first gift, possibly with a sequence of ropes and another gifts.

Output

If it is possible for Chloe and Vladik to choose prizes without fighting, print single integer — the maximum possible sum of pleasantness they can get together.

Otherwise print Impossible.

Examples

input

8
0 5 -1 4 3 2 6 5
1 2
2 4
2 5
1 3
3 6
6 7
6 8

output

25

input

4
1 -5 1 1
1 2
1 4
2 3

output

2

input

1
-1

output

Impossible

题意：给一棵树，1是根，每个结点都有一个权值，求两棵不交叉子树的最大权值。

题解：DFS下，mx[u]维护一下以u为根的子树的最大权值。

#include <bits/stdc++.h>
using namespace std;

#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define pb push_back
#define mp make_pair
#define cl(a) memset((a),0,sizeof(a))
#ifdef HandsomeHow
#define dbg(x) cerr << #x << " = " << x << endl
#else
#define dbg(x)
#endif
typedef long long ll;
typedef unsigned long long ull;
typedef pair <int, int> pii;
const int inf=0x3f3f3f3f;
const double eps=1e-8;
const int mod=1000000007;
const double pi=acos(-1.0);
inline void gn(long long&x){
int sg=1;char c;while(((c=getchar())<'0'||c>'9')&&c!='-');c=='-'?(sg=-1,x=0):(x=c-'0');
while((c=getchar())>='0'&&c<='9')x=x*10+c-'0';x*=sg;}
inline void gn(int&x){long long t;gn(t);x=t;}
inline void gn(unsigned long long&x){long long t;gn(t);x=t;}
ll gcd(ll a,ll b){return a? gcd(b%a,a):b;}
ll powmod(ll a,ll x,ll mod){ll t=1ll;while(x){if(x&1)t=t*a%mod;a=a*a%mod;x>>=1;}return t;}
// (づ°ω°)づe★------------------------------------------------
const int maxn = 2e5+5;
vector<int>E[maxn];
ll su[maxn];
ll mx[maxn];
ll ans;
ll v[maxn];

void dfs(int u, int p){
su[u] = v[u];
mx[u] = -18000000000000000ll;
for(auto it:E[u]){
if(it == p) continue;
dfs(it,u);
mx[u] = max(mx[u],mx[it]);
su[u]+=su[it];
}
mx[u] = max(mx[u],su[u]);
}

void dfs2(int u, int p){
for(auto it:E[u]){
if(it == p) continue;
dfs2(it,u);
}
if((u == 1 && E[1].size()>=2) || (u!=1 && E[u].size()>2)){
vector<ll>tmp;
for(auto it:E[u]){
if(it == p) continue;
tmp.pb(mx[it]);
}
sort(tmp.begin(),tmp.end());
int sz = tmp.size();
ans = max(ans,tmp[sz-1]+tmp[sz-2]);
}
}

int main(){
#ifdef HandsomeHow
freopen("data.in","r",stdin);
//freopen("data.out","w",stdout);
#endif
int n;
gn(n);
rep(i,1,n) gn(v[i]);
rep(i,2,n){
int x,y;
gn(x);gn(y);
E[x].pb(y);E[y].pb(x);
}
dfs(1,-1);
ans = -18000000000000000ll;
dfs2(1,-1);
if(ans == -18000000000000000ll)
puts("Impossible");
else cout<<ans<<endl;
return 0;
}