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 double PI = acos(-1.0);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int rnd(int l, int r) { return l + rng() % (r - l + 1); }
namespace IO {
bool REOF = 1;
inline char nc() {
  static char buf[100000], *p1 = buf, *p2 = buf;
  return p1 == p2 && REOF &&
                 (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? (REOF = 0, EOF)
             : *p1++;
}
template <class T>
inline bool read(T &first) {
  char c = nc();
  bool f = 0;
  first = 0;
  while (c < '0' || c > '9') c == '-' && (f = 1), c = nc();
  while (c >= '0' && c <= '9')
    first = (first << 3) + (first << 1) + (c ^ 48), c = nc();
  if (f) first = -first;
  return REOF;
}
template <typename T, typename... T2>
inline bool read(T &first, T2 &...rest) {
  read(first);
  return read(rest...);
}
inline bool need(char &c) {
  return ((c >= 'a') && (c <= 'z')) || ((c >= '0') && (c <= '9')) ||
         ((c >= 'A') && (c <= 'Z'));
}
inline bool read_str(char *a) {
  while ((*a = nc()) && need(*a) && REOF) ++a;
  *a = '\0';
  return REOF;
}
inline bool read_dbl(double &first) {
  bool f = 0;
  char ch = nc();
  first = 0;
  while (ch < '0' || ch > '9') {
    f |= (ch == '-');
    ch = nc();
  }
  while (ch >= '0' && ch <= '9') {
    first = first * 10.0 + (ch ^ 48);
    ch = nc();
  }
  if (ch == '.') {
    double tmp = 1;
    ch = nc();
    while (ch >= '0' && ch <= '9') {
      tmp = tmp / 10.0;
      first = first + tmp * (ch ^ 48);
      ch = nc();
    }
  }
  if (f) first = -first;
  return REOF;
}
template <class TH>
void _dbg(const char *sdbg, TH h) {
  cerr << sdbg << '=' << h << '\n';
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << '=' << h << ',' << ' ';
  _dbg(sdbg + 1, a...);
}
template <class T>
ostream &operator<<(ostream &os, vector<T> V) {
  os << "[";
  for (auto vv : V) os << vv << ",";
  return os << "]";
}
template <class T>
ostream &operator<<(ostream &os, set<T> V) {
  os << "[";
  for (auto vv : V) os << vv << ",";
  return os << "]";
}
template <class T>
ostream &operator<<(ostream &os, map<T, T> V) {
  os << "[";
  for (auto vv : V) os << vv << ",";
  return os << "]";
}
template <class L, class R>
ostream &operator<<(ostream &os, pair<L, R> P) {
  return os << "(" << P.st << "," << P.nd << ")";
}
}  // namespace IO
using namespace IO;
const int maxn = 2e5 + 5;
const int maxv = 2e5 + 5;
const int mod = 998244353;
const int INF = 1e9 + 7;
const double eps = 1e-12;
int dx[4] = {0, 1, 0, -1};
int dy[4] = {1, 0, -1, 0};
int tot, head[maxn];
struct Edge {
  int v, nxt;
  Edge() {}
  Edge(int _v, int _nxt) : v(_v), nxt(_nxt) {}
} e[maxn << 1];
void init() {
  tot = 1;
  memset(head, 0, sizeof(head));
}
void addedge(int u, int v) {
  e[tot] = Edge(v, head[u]);
  head[u] = tot++;
  e[tot] = Edge(u, head[v]);
  head[v] = tot++;
}
void addarc(int u, int v) {
  e[tot] = Edge(v, head[u]);
  head[u] = tot++;
}
int n, a[maxn], ans[maxn], val[maxn];
void dfs1(int u, int f) {
  val[u] = a[u];
  for (int i = head[u]; i; i = e[i].nxt) {
    int v = e[i].v;
    if (v == f) continue;
    dfs1(v, u);
    if (val[v] >= 0) val[u] += val[v];
  }
}
void dfs2(int u, int f) {
  ans[u] = val[u];
  if (val[u] > 0) ans[f] -= val[u];
  if (ans[f] > 0) ans[u] += ans[f];
  for (int i = head[u]; i; i = e[i].nxt) {
    int v = e[i].v;
    if (v == f) continue;
    dfs2(v, u);
  }
  if (val[u] > 0) ans[f] += val[u];
}
void solve() {
  read(n);
  for (int i = 1; i <= n; i++) {
    read(a[i]);
    if (a[i] == 0) a[i] = -1;
  }
  int u, v;
  init();
  for (int i = 1; i <= n - 1; i++) {
    read(u, v);
    addedge(u, v);
  }
  dfs1(1, 0);
  dfs2(1, 0);
  for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
}
int main() {
  solve();
  return 0;
}