Problem: You are given a tree consisting of n vertices. A tree is a connected undirected graph with n-1 edges. Each vertex v of this tree has a color assigned to it (a_v = 1 if the vertex v is white and 0 if the vertex v is black).

You have to solve the following problem for each vertex v: what is the maximum difference between the number of white and the number of black vertices you can obtain if you choose some subtree of the given tree that contains the vertex v? The subtree of the tree is the connected subgraph of the given tree. More formally, if you choose the subtree that contains cnt_w white vertices and cnt_b black vertices, you have to maximize cnt_w - cnt_b.

Input

The first line of the input contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of vertices in the tree.

The second line of the input contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1), where a_i is the color of the i-th vertex.

Each of the next n-1 lines describes an edge of the tree. Edge i is denoted by two integers u_i and v_i, the labels of vertices it connects (1 ≤ u_i, v_i ≤ n, u_i ≠ v_i).

It is guaranteed that the given edges form a tree.

Output

Print n integers res_1, res_2, ..., res_n, where res_i is the maximum possible difference between the number of white and black vertices in some subtree that contains the vertex i.

Examples

Input


9
0 1 1 1 0 0 0 0 1
1 2
1 3
3 4
3 5
2 6
4 7
6 8
5 9


Output


2 2 2 2 2 1 1 0 2 


Input


4
0 0 1 0
1 2
1 3
1 4


Output


0 -1 1 -1 

Note

The first example is shown below:

<image>

The black vertices have bold borders.

In the second example, the best subtree for vertices 2, 3 and 4 are vertices 2, 3 and 4 correspondingly. And the best subtree for the vertex 1 is the subtree consisting of vertices 1 and 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace ::std;
const long double PI = acos(-1);
const long long MOD = 1000000000 + 7;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long add(long long a, long long b, long long m = MOD) {
  if (a >= m) a %= m;
  if (b >= m) b %= m;
  if (a < 0) a += m;
  if (b < 0) b += m;
  long long res = a + b;
  if (res >= m or res <= -m) res %= m;
  if (res < 0) res += m;
  return res;
}
long long mul(long long a, long long b, long long m = MOD) {
  if (a >= m) a %= m;
  if (b >= m) b %= m;
  if (a < 0) a += m;
  if (b < 0) b += m;
  long long res = a * b;
  if (res >= m or res <= -m) res %= m;
  if (res < 0) res += m;
  return res;
}
long long pow_mod(long long a, long long b, long long m = MOD) {
  long long res = 1LL;
  a = a % m;
  while (b) {
    if (b & 1) res = mul(res, a, m);
    b >>= 1;
    a = mul(a, a, m);
  }
  return res;
}
long long fastexp(long long a, long long b) {
  long long res = 1LL;
  while (b) {
    if (b & 1) res = res * a;
    b >>= 1;
    a *= a;
  }
  return res;
}
int gcdExtendido(int a, int b, int *x, int *y) {
  if (a == 0) {
    *x = 0;
    *y = 1;
    return b;
  }
  int x1, y1;
  int gcd = gcdExtendido(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
int modInverso(int a, int m) {
  int x, y;
  int g = gcdExtendido(a, m, &x, &y);
  if (g != 1)
    return -1;
  else
    return (x % m + m) % m;
}
const int N = 200000 + 5;
int n;
vector<int> G[N];
int a[N];
int in[N];
int out[N];
int memo[N];
void DFS(int u, int id, int p = -1) {
  if (id == 0) {
    in[u] = a[u];
    for (int v : G[u]) {
      if (v == p) continue;
      DFS(v, id, u);
      in[u] += max(0, in[v]);
    }
  } else {
    int bestsum = a[u];
    for (int v : G[u]) {
      if (v == p) continue;
      bestsum += max(0, in[v]);
    }
    for (int v : G[u]) {
      if (v == p) continue;
      int bestadd = bestsum + out[u] - a[u];
      if (in[v] >= 0) bestadd -= in[v];
      bestadd = max(bestadd, 0);
      out[v] = bestadd + a[v];
    }
    for (int v : G[u]) {
      if (v == p) continue;
      DFS(v, id, u);
    }
  }
}
void preprocess() {
  for (int i = 1; i <= n; i++) {
    memo[i] = in[i] + out[i] - a[i];
  }
}
int main() {
  scanf("%d", &(n));
  for (int i = 1; i <= n; i++) {
    scanf("%d", &(a[i]));
    if (a[i] == 0) a[i] = -1;
  }
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d %d", &(u), &(v));
    G[u].emplace_back(v);
    G[v].emplace_back(u);
  }
  DFS(1, 0);
  out[1] = a[1];
  DFS(1, 1);
  preprocess();
  for (int i = 1; i <= n; i++) {
    printf("%d%c", memo[i], " \n"[i == n]);
  }
  return 0;
}