#include <bits/stdc++.h>

int ri() {
	int n;
	scanf("%d", &n);
	return n;
}

std::vector<int> color;
std::vector<std::vector<int> > hen;

std::vector<int> size;
void dfs1(int i, int prev) {
	for (auto j : hen[i]) if (j != prev) dfs1(j, i), size[i] += size[j];
}

std::vector<int> cur;
std::vector<int64_t> res;
void dfs2(int i, int prev) {
	int bak = cur[color[i]];
	for (auto j : hen[i]) if (j != prev) {
		cur[color[i]] = size[j];
		dfs2(j, i);
		res[color[i]] += (int64_t) size[j] * (size[j] + 1) / 2;
	}
	cur[color[i]] = bak - size[i];
	res[color[i]] -= (int64_t) bak * (bak + 1) / 2;
	res[color[i]] += (int64_t) cur[color[i]] * (cur[color[i]] + 1) / 2;
}

int main() {
	int n = ri();
	color.resize(n);
	for (auto &i : color) i = ri() - 1;
	hen.resize(n);
	for (int i = 1; i < n; i++) {
		int a = ri() - 1;
		int b = ri() - 1;
		hen[a].push_back(b);
		hen[b].push_back(a);
	}
	size.resize(n, 1);
	dfs1(0, -1);
	cur.resize(n, size[0]);
	res.resize(n, (int64_t) n * (n + 1) / 2);
	dfs2(0, -1);
	for (auto i : res) printf("%" PRId64 "\n", (int64_t) n * (n + 1) / 2 - i);
	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>

int ri() {
	int n;
	scanf("%d", &n);
	return n;
}

std::vector<int> color;
std::vector<std::vector<int> > hen;

std::vector<int> size;
void dfs1(int i, int prev) {
	for (auto j : hen[i]) if (j != prev) dfs1(j, i), size[i] += size[j];
}

std::vector<int> cur;
std::vector<int64_t> res;
void dfs2(int i, int prev) {
	int bak = cur[color[i]];
	for (auto j : hen[i]) if (j != prev) {
		cur[color[i]] = size[j];
		dfs2(j, i);
		res[color[i]] += (int64_t) size[j] * (size[j] + 1) / 2;
	}
	cur[color[i]] = bak - size[i];
	res[color[i]] -= (int64_t) bak * (bak + 1) / 2;
	res[color[i]] += (int64_t) cur[color[i]] * (cur[color[i]] + 1) / 2;
}

int main() {
	int n = ri();
	color.resize(n);
	for (auto &i : color) i = ri() - 1;
	hen.resize(n);
	for (int i = 1; i < n; i++) {
		int a = ri() - 1;
		int b = ri() - 1;
		hen[a].push_back(b);
		hen[b].push_back(a);
	}
	size.resize(n, 1);
	dfs1(0, -1);
	cur.resize(n, size[0]);
	res.resize(n, (int64_t) n * (n + 1) / 2);
	dfs2(0, -1);
	for (auto i : res) printf("%" PRId64 "\n", (int64_t) n * (n + 1) / 2 - i);
	return 0;
}

```