codeforce 743 D. Chloe and pleasant prizes (树形dp)

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

第一道树形dp的题目；
思路:该题目已经告诉我们所有的点都会和1相连,也就是说此树的根为1,我一开始还想从叶节点到根节点去找的,
对于每个以该点为子树的节点 他都有两种可能的取值 一个是dp[i]，一个为 num[i] 。dp[i]  表示以该点为根的子树当中权值
最大的,  num[i]表示以 该点为根的子树 的所有权值（包括自己）;
我们要想使两个子树的和最大,那么就是在以该点为根的子树当中找到两个 和最大的即可

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
const int N = 2e5 + 100;
vector<int>vec
;
ll a
,sum
,dp
,ans;
int vis
;
const ll inf = (ll)1e18 + 1;
void dfs(int rt)
{
ll Max=-inf,Min=-inf;
int i,j,son;
sum[rt]=a[rt];
for(i=0;i<vec[rt].size();i++) {
son=vec[rt][i];
if(vis[son]) continue;
vis[son]=1;
dfs(son);
sum[rt]+=sum[son];
if(sum[son]>dp[son]) dp[son]=sum[son];
if(dp[son]>dp[rt]) Min=Max,Max=dp[son],dp[rt]=Max;
else Min=max(Min,dp[son]);
if(Min!=-inf ) ans=max(ans,Max+Min);
}
}

int main()
{
ios::sync_with_stdio(false);
int i,j,n;
int u,v;
cin>>n;
for(i=1;i<=n;i++) {
cin>>a[i];
dp[i]=-inf;
}
for(i=1;i<n;i++) {
cin>>u>>v;
vec[u].push_back(v);
vec[v].push_back(u);
}
ans=-inf;
vis[1]=1;
dfs(1);
if(ans==-inf) cout<<"Impossible"<<endl;
else cout<<ans<<endl;
return 0;
}