【27.85%】【codeforces 743D】Chloe and pleasant prizes

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard 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

【题目链接】:http://codeforces.com/contest/743/problem/D

【题解】



假设现在搞到第v个节点;

则这两个子树只能在v的两个不同的儿子节点(或它的儿子节点..)中选两个(当然不一定是最终答案但可能是);

看看选哪两个最优即可;

设f[x]表示x节点以下(包括x节点)的子树中子树和最大的子树的和.

在枚举儿子i的过程中维护前i-1个儿子的f[x]的最大值就好;

和当前的儿子i相加看看能不能让答案更优.

(找到第一第二大的和的方法);

如果选了x节点。

则x节点以下的所有节点都会被选.

所以维护一下以x节点为根节点的子树的和sum[x];



【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int MAXN = 2e5+100;
const LL INF = 1e15;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);

LL a[MAXN],f[MAXN],ans = -INF,sum[MAXN];
vector <int> G[MAXN];
int n;

void dfs(int x,int fa)
{
    int len = G[x].size();
    LL temp = a[x];
    rep1(i,0,len-1)
    {
        int y = G[x][i];
        if (y==fa)
            continue;
        dfs(y,x);
        temp+=sum[y];
        if (f[x]!=-INF)
            ans = max(ans,f[x]+f[y]);
        f[x] = max(f[x],f[y]);
    }
    sum[x] = temp;
    f[x] = max(f[x],sum[x]);
}

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    rei(n);
    rep1(i,1,n)
        rel(a[i]);
    rep1(i,1,n-1)
    {
        int x,y;
        rei(x);rei(y);
        G[x].pb(y);
        G[y].pb(x);
    }
    rep1(i,1,n)
        f[i] = -INF;
    dfs(1,-1);
    if (ans==-INF)
        puts("Impossible");
    else
        cout << ans << endl;
    return 0;
}