Tree with Maximum Cost（樹形DP）

 

You are given a tree consisting exactly of nn vertices. Tree is a connected undirected graph with n−1n−1 edges. Each vertex vv of this tree has a value avav assigned to it.

Let dist(x,y)dist(x,y) be the distance between the vertices xx and yy. 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 vv. Then the cost of the tree is ∑i=1ndist(i,v)⋅ai∑i=1ndist(i,v)⋅ai.

Your task is to calculate the maximum possible cost of the tree if you can choose vvarbitrarily.

Input

The first line contains one integer nn, the number of vertices in the tree (1≤n≤2⋅1051≤n≤2⋅105).

The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤2⋅1051≤ai≤2⋅105), where aiai is the value of the vertex ii.

Each of the next n−1n−1 lines describes an edge of the tree. Edge ii is denoted by two integers uiui and vivi, the labels of vertices it connects (1≤ui,vi≤n1≤ui,vi≤n, ui≠viui≠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 vv.

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

給定一棵樹 ，u點關於數的權值是∑distance (i, u）×value ( i ),求得所有點中對應數的全值的最大值

sum[ i ]爲 i 和 i 的子樹的權值和；dp[ i ] 爲以i點爲根節點的樹所求出的這個子樹的權值，

dp[1]爲當u等於1時的值，則，dp[2] = dp[1] -sum[2]+sum[1]-sum[2]

因此dp[i] = dp[father]  + sum[1]- 2*dp[i]

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

struct node
{
    int to,next;
} edge[400005];

long long head[200005],dp[200005],sum[200005],cnt,ans;

void add(int u, int v)
{
    edge[cnt].to = v;
    edge[cnt].next = head[u];
    head[u] = cnt++;
}


void bfs(int u, int father)
{
    for(int i = head[u]; i != -1; i = edge[i].next)
    {
        int v = edge[i].to;
        if(v == father)
            continue;
        bfs(v,u);
        sum[u] += sum[v];
        dp[u] += dp[v] + sum[v];
    }
}


void bfs2(int u, int father)
{
    if(u !=1)
        dp[u] = dp[father] + sum[1]-2*sum[u];
    ans = max(ans,dp[u]);
    for(int i = head[u]; i != -1; i = edge[i].next)
    {
        int v = edge[i].to;
        if(v == father)
            continue;
        bfs2(v,u);

    }
    return ;
}

int main()
{
    int i,n,v,u;
    scanf("%d",&n);
    memset(head,-1,sizeof(head));
    memset(dp,0,sizeof(dp));
    for(i = 1; i <= n; i ++)
    {
        scanf("%lld",&sum[i]);
    }
    cnt = 0;
    for(i = 0; i < n-1; i ++)
    {
        scanf("%d %d",&u, &v);
        add(u,v);
        add(v,u);
    }
    bfs(1,0);
    bfs2(1,0);
    printf("%lld\n",ans);

    return 0;
}