#include <bits/stdc++.h>
#define debug(...) fprintf(stderr, __VA_ARGS__)
using namespace std;

void read (int&a) {
	a = 0; char k = getchar(); int f = 1;
	for(;!isdigit(k); k = getchar()) if (k == '-') f = -1;
	for(; isdigit(k); k = getchar()) a = a * 10 + k - '0';
	a *= f;
}

const int N = 200000 + 5;
vector<int> G[N];
int n, c[N], sz[N];
long long T[N], ans[N];

void dfs (int u, int f) {
	long long tmp = T[c[u]]; T[c[u]] = 0;
	sz[u] = 1;
	for (int i = 0; i < (signed)G[u].size(); ++i) if (G[u][i] != f) {
		int v = G[u][i];
		dfs (v, u);
		ans[c[u]] -= 1LL * (sz[v] - T[c[u]]) * ((sz[v] - T[c[u]]) + 1) / 2;
		sz[u] += sz[v];
		T[c[u]] = 0;
	}
	T[c[u]] = tmp + sz[u];
}

int main ()
{
	read (n);
	for (int i = 1; i <= n; ++i) read (c[i]);
	for (int i = 1; i < n; ++i) {
		int x, y; read (x), read (y);
		G[x].push_back (y),
		G[y].push_back (x);
	}
	dfs (1, 0);
	long long alter = 1LL * n * (n + 1) / 2;
	for (int i = 1; i <= n; ++i) printf ("%lld\n", alter + ans[i] - 1LL * (n - T[i]) * (n - T[i] + 1) / 2);
	return 0;
}
















### Prompt

Your challenge is to write a CPP solution to the following problem:

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>
#define debug(...) fprintf(stderr, __VA_ARGS__)
using namespace std;

void read (int&a) {
	a = 0; char k = getchar(); int f = 1;
	for(;!isdigit(k); k = getchar()) if (k == '-') f = -1;
	for(; isdigit(k); k = getchar()) a = a * 10 + k - '0';
	a *= f;
}

const int N = 200000 + 5;
vector<int> G[N];
int n, c[N], sz[N];
long long T[N], ans[N];

void dfs (int u, int f) {
	long long tmp = T[c[u]]; T[c[u]] = 0;
	sz[u] = 1;
	for (int i = 0; i < (signed)G[u].size(); ++i) if (G[u][i] != f) {
		int v = G[u][i];
		dfs (v, u);
		ans[c[u]] -= 1LL * (sz[v] - T[c[u]]) * ((sz[v] - T[c[u]]) + 1) / 2;
		sz[u] += sz[v];
		T[c[u]] = 0;
	}
	T[c[u]] = tmp + sz[u];
}

int main ()
{
	read (n);
	for (int i = 1; i <= n; ++i) read (c[i]);
	for (int i = 1; i < n; ++i) {
		int x, y; read (x), read (y);
		G[x].push_back (y),
		G[y].push_back (x);
	}
	dfs (1, 0);
	long long alter = 1LL * n * (n + 1) / 2;
	for (int i = 1; i <= n; ++i) printf ("%lld\n", alter + ans[i] - 1LL * (n - T[i]) * (n - T[i] + 1) / 2);
	return 0;
}
















```