Codeforces Round #527 (Div. 3) Tree with Maximum Cost

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 avavassigned 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

Note

Picture corresponding to the first example: 

You can choose the vertex 33 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=1212⋅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 00.

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<queue>
#include<string>
#include<map>
#include<cmath>
#define INF 0x3f3f3f3f
#define rep(i,a,n) for(int i=a;i<n;i++)
#define per(i,a,n) for(int i=n-1;i>=a;i--)
#define fori(x) for(int i=0;i<x;i++)
#define forj(x) for(int j=0;j<x;j++)
#define memset(x,y) memset(x,y,sizeof(x))
#define memcpy(x,y) memcpy(x,y,sizeof(y))
#define sca(x) scanf("%d", &x)
#define scas(x) scanf("%s",x)
#define sca2(x,y) scanf("%d%d",&x,&y)
#define sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define scl(x) scanf("%lld",&x)
#define scl2(x,y) scanf("%lld%lld",&x,&y)
#define scl3(x,y,z) scanf("%lld%lld%lld",&x,&y,&z)
#define pri(x) printf("%d\n",x)
#define pri2(x,y) printf("%d %d\n",x,y)
#define pris(x) printf("%s\n",x)
#define prl(x) printf("%lld\n",x)
//#include <bits/stdc++.h>

typedef long long ll;
const int maxn=2e5+10;
const int mod=1e9+7;
const double eps=1e-8;
// const double pi = acos(-1);

using namespace std;

ll dp[maxn];
int a[maxn];
ll sum;
ll val[maxn];
vector<int> v[maxn];
ll res;
void dfs(int u,int f)
{
    val[u] = a[u];
    rep(i,0,v[u].size())
    {
      int nex = v[u][i];
      if(nex == f)
        continue;
      dfs(nex,u);
      dp[u] += dp[nex]+val[nex];
      val[u] += val[nex];

    }
}
void ddfs(int u,int f)
{
  res = max(res,dp[u]);
  rep(i,0,v[u].size())
  {
    int nex = v[u][i];
    if(nex == f)
      continue;
    dp[nex] = dp[u] - val[nex] + (sum - val[nex]);
    ddfs(nex,u);
  }
}
int main()
{
    int n;
    sca(n);
    sum = 0;
    rep(i,1,n+1)
    {
      sca(a[i]);
      sum += a[i];
    }
    rep(i,0,n-1)
    {
      int x,y;
      sca2(x,y);
      v[x].push_back(y);
      v[y].push_back(x);
    }
    res = -1;

    dfs(1,0);
    ddfs(1,0);
    prl(res);
    return 0;
}