#include<bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); ++i)
#define rep2(i, s, n) for (int i = (s); i < (int)(n); ++i)
#define chmax(a, b) a = max(a, b)
#define chmin(a, b) a = min(a, b)
#define fi first
#define se second
#define INF 1000000009
#define LLINF 1000000000000000009LL
using ll = long long;

const int MX = 200009;
vector<int> to[MX];
int t[MX];
int c[MX];
ll ans[MX];
ll f(ll n) { return n * (n + 1) / 2; }

int tfs(int v, int p=-1) {
  t[v] = 1;
  for (auto u : to[v]) {
    if (u == p) continue;
    t[v] += tfs(u, v);
  }
  return t[v];
}

stack<int> st[MX];
int d[MX], remsz[MX];
void dfs(int v, int p=-1) {
  st[c[v]].push(v);
  for (int u : to[v]) {
    if (u == p) continue;
    d[v] = t[u];
    dfs(u, v);
    ans[c[v]] -= f(d[v]);
  }
  st[c[v]].pop();
  if (st[c[v]].size()) {
    d[st[c[v]].top()] -= t[v];
  }
  else {
    remsz[c[v]] -= t[v];
  }
}

int main() {
  int n;
  cin >> n;
  rep(i,n) cin >> c[i], c[i]--;
  rep(i,n-1) {
    int a, b;
    cin >> a >> b;
    a--; b--;
    to[a].push_back(b);
    to[b].push_back(a);
  }
  tfs(0);
  rep(i,n) ans[i] = f(n);
  rep(i,n) remsz[i] = n;
  dfs(0);
  rep(i,n) ans[i] -= f(remsz[i]);
  rep(i,n) cout << ans[i] << endl;
  return(0);
}

### Prompt

Please provide a cpp coded solution to the problem described below:

We have a tree with N vertices numbered 1 to N. The i-th edge in this tree connects Vertex a_i and b_i. Additionally, each vertex is painted in a color, and the color of Vertex i is c_i. Here, the color of each vertex is represented by an integer between 1 and N (inclusive). The same integer corresponds to the same color; different integers correspond to different colors.

For each k=1, 2, ..., N, solve the following problem:

* Find the number of simple paths that visit a vertex painted in the color k one or more times.



Note: The simple paths from Vertex u to v and from v to u are not distinguished.

Constraints

* 1 \leq N \leq 2 \times 10^5
* 1 \leq c_i \leq N
* 1 \leq a_i,b_i \leq N
* The given graph is a tree.
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
c_1 c_2 ... c_N
a_1 b_1
:
a_{N-1} b_{N-1}


Output

Print the answers for k = 1, 2, ..., N in order, each in its own line.

Examples

Input

3
1 2 1
1 2
2 3


Output

5
4
0


Input

1
1


Output

1


Input

2
1 2
1 2


Output

2
2


Input

5
1 2 3 4 5
1 2
2 3
3 4
3 5


Output

5
8
10
5
5


Input

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


Output

18
15
0
14
23
0
23
0

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); ++i)
#define rep2(i, s, n) for (int i = (s); i < (int)(n); ++i)
#define chmax(a, b) a = max(a, b)
#define chmin(a, b) a = min(a, b)
#define fi first
#define se second
#define INF 1000000009
#define LLINF 1000000000000000009LL
using ll = long long;

const int MX = 200009;
vector<int> to[MX];
int t[MX];
int c[MX];
ll ans[MX];
ll f(ll n) { return n * (n + 1) / 2; }

int tfs(int v, int p=-1) {
  t[v] = 1;
  for (auto u : to[v]) {
    if (u == p) continue;
    t[v] += tfs(u, v);
  }
  return t[v];
}

stack<int> st[MX];
int d[MX], remsz[MX];
void dfs(int v, int p=-1) {
  st[c[v]].push(v);
  for (int u : to[v]) {
    if (u == p) continue;
    d[v] = t[u];
    dfs(u, v);
    ans[c[v]] -= f(d[v]);
  }
  st[c[v]].pop();
  if (st[c[v]].size()) {
    d[st[c[v]].top()] -= t[v];
  }
  else {
    remsz[c[v]] -= t[v];
  }
}

int main() {
  int n;
  cin >> n;
  rep(i,n) cin >> c[i], c[i]--;
  rep(i,n-1) {
    int a, b;
    cin >> a >> b;
    a--; b--;
    to[a].push_back(b);
    to[b].push_back(a);
  }
  tfs(0);
  rep(i,n) ans[i] = f(n);
  rep(i,n) remsz[i] = n;
  dfs(0);
  rep(i,n) ans[i] -= f(remsz[i]);
  rep(i,n) cout << ans[i] << endl;
  return(0);
}

```