You are given a tree consisting exactly of n vertices. Tree is a connected undirected graph with n−1 edges. Each vertex v of this tree has a value av assigned to it.
Let dist(x,y) be the distance between the vertices x and y. The distance between the vertices is the number of edges on the simple path between them.
Let’s define the cost of the tree as the following value: firstly, let’s fix some vertex of the tree. Let it be v. Then the cost of the tree is ∑i=1ndist(i,v)⋅ai.
Your task is to calculate the maximum possible cost of the tree if you can choose v arbitrarily.
Input
The first line contains one integer n, the number of vertices in the tree (1≤n≤2⋅105).
The second line of the input contains n integers a1,a2,…,an (1≤ai≤2⋅105), where ai is the value of the vertex i.
Each of the next n−1 lines describes an edge of the tree. Edge i is denoted by two integers ui and vi, the labels of vertices it connects (1≤ui,vi≤n, ui≠vi).
It is guaranteed that the given edges form a tree.
Output
Print one integer — the maximum possible cost of the tree if you can choose any vertex as v.
Examples
Input
8
9 4 1 7 10 1 6 5
1 2
2 3
1 4
1 5
5 6
5 7
5 8
Output
121
Input
1
1337
Output
0
Note
Picture corresponding to the first example:
You can choose the vertex 3 as a root, then the answer will be 2⋅9+1⋅4+0⋅1+3⋅7+3⋅10+4⋅1+4⋅6+4⋅5=18+4+0+21+30+4+24+20=121.
In the second example tree consists only of one vertex so the answer is always 0.

放一个跟别人思路一样，过程稍有不同的代码。
具体是转移方程不好想，求出其中一个点为根的权值，向其子节点转移，
公式：dp[next]
=dp[now]-要转移的方向子树权值和+剩下子树权值和
=dp[now]-要转移的方向子树权值和+（all-要转移的方向子树权值和）

#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <queue>
#include <vector>
#include <map>
#include <string>

using namespace std;
//for(i=1;i<n;i++)
//scanf("%d",&n);
//printf("\n",);
long long dp[300010],money[300010],temp[300010],ans,all;

vector<int>mapp[300010];

int dfs(int point,int step,int now)
{
    dp[1]+=now*money[point];
    temp[point]=money[point];
    for(int i=0;i<mapp[point].size();i++)
    {
        if(step==mapp[point][i])
            continue;

        dfs(mapp[point][i],point,now+1);
        temp[point]+=temp[mapp[point][i]];

    }
    if(mapp[point].size()==1)
    temp[point]=money[point];

    return 0;
}

int ddffss(int point,int step)
{
    ans=max(ans,dp[point]);
    for(int i=0;i<mapp[point].size();i++)
    {
        if(step==mapp[point][i])
            continue;

        dp[mapp[point][i]]=dp[point]-temp[mapp[point][i]]*2+all;
        ddffss(mapp[point][i],point);

    }

    return 0;
}


int main()
{
    int i,j,l,a,b,k;
    all=0;


    scanf("%d",&a);

    for(i=1;i<=a;i++)
    {
        scanf("%lld",&money[i]);
        all+=money[i];
    }


    for(i=1;i<a;i++)
    {
        scanf("%d%d",&j,&k);
        mapp[j].push_back(k);
        mapp[k].push_back(j);
    }


    dfs(1,-1,0);
    ans=dp[1];
    ddffss(1,-1);

    printf("%lld",ans);

    return 0;
}