#include<bits/stdc++.h>
using namespace std;

const int N = 2e5 + 5;
typedef long long ll;

int n, a[N], tot, first[N], fa[N], top[N], up[N], d[N];
ll ans[N], cnt[N], s[N];
struct Edge {
	int to, next;
} e[N<<1];

void Dfs2(int u) {
	up[u] = top[a[u]];
	ll out;
	if(up[u]) {
		cnt[up[u]] -= s[u];
		out = cnt[up[u]];
	}
	else { 
	    d[a[u]] -= s[u];
	    out = d[a[u]]; 
	}
	ans[a[u]] += s[u]*(s[u] - 1)/2 + 1 + out*s[u];
	top[a[u]] = u;
	for(int i = first[u]; i; i = e[i].next) {
		int v = e[i].to;
		if(v == fa[u]) continue;
		ans[a[u]] -= s[v]*(s[v] - 1)/2;
		cnt[u] = s[v];
		Dfs2(v);
	}
	top[a[u]] = up[u];
}

void Dfs1(int u, int f) {
	s[u] = 1;
	fa[u] = f;
	for(int i = first[u]; i; i = e[i].next) {
		int v = e[i].to;
		if(v == fa[u]) continue;
		Dfs1(v, u);
		s[u] += s[v];
	}
	cnt[u] = s[u] - 1;
}

void add(int u, int v) { 
	e[++tot] = (Edge) {v, first[u]};
	first[u] = tot;
}

int main() {
	cin>>n;
	for(int i = 1; i <= n; i++)
		cin>>a[i], d[i] = n;
	for(int u, v, i = 1; i < n; i++) {
		cin>>u>>v;
		add(u, v);
		add(v, u);
	}
	Dfs1(1, -1);
    s[0] = n;
    Dfs2(1);
	for(int i = 1; i <= n; i++)
		cout<<ans[i]<<endl;
	return 0;
}
### Prompt

In CPP, your task is to solve 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>
using namespace std;

const int N = 2e5 + 5;
typedef long long ll;

int n, a[N], tot, first[N], fa[N], top[N], up[N], d[N];
ll ans[N], cnt[N], s[N];
struct Edge {
	int to, next;
} e[N<<1];

void Dfs2(int u) {
	up[u] = top[a[u]];
	ll out;
	if(up[u]) {
		cnt[up[u]] -= s[u];
		out = cnt[up[u]];
	}
	else { 
	    d[a[u]] -= s[u];
	    out = d[a[u]]; 
	}
	ans[a[u]] += s[u]*(s[u] - 1)/2 + 1 + out*s[u];
	top[a[u]] = u;
	for(int i = first[u]; i; i = e[i].next) {
		int v = e[i].to;
		if(v == fa[u]) continue;
		ans[a[u]] -= s[v]*(s[v] - 1)/2;
		cnt[u] = s[v];
		Dfs2(v);
	}
	top[a[u]] = up[u];
}

void Dfs1(int u, int f) {
	s[u] = 1;
	fa[u] = f;
	for(int i = first[u]; i; i = e[i].next) {
		int v = e[i].to;
		if(v == fa[u]) continue;
		Dfs1(v, u);
		s[u] += s[v];
	}
	cnt[u] = s[u] - 1;
}

void add(int u, int v) { 
	e[++tot] = (Edge) {v, first[u]};
	first[u] = tot;
}

int main() {
	cin>>n;
	for(int i = 1; i <= n; i++)
		cin>>a[i], d[i] = n;
	for(int u, v, i = 1; i < n; i++) {
		cin>>u>>v;
		add(u, v);
		add(v, u);
	}
	Dfs1(1, -1);
    s[0] = n;
    Dfs2(1);
	for(int i = 1; i <= n; i++)
		cout<<ans[i]<<endl;
	return 0;
}
```