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

#define ll long long
#define mp make_pair
#define pb push_back

#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#define rep(i, n) for (int i = 0; i < (int)(n); ++ i)

const int mxn = 2e5 + 5;
int n, c[mxn], hd[mxn], nex[mxn << 1], to[mxn << 1], tot, s[mxn], sz[mxn];
ll ans[mxn];

void add_edge(int u, int v) { nex[++ tot] = hd[u]; hd[u] = tot; to[tot] = v; }

void dfs(int u, int p) {
	sz[u] = 1;
	for (int e = hd[u]; e; e = nex[e]) {
		int v = to[e];
		if (v == p) continue;
		int old = s[c[u]];
		dfs(v, u);
		int nw = s[c[u]];
		int f = sz[v] - (nw - old);
		ans[c[u]] -= 1LL * f * (f + 1) / 2;
		s[c[u]] += f;
		sz[u] += sz[v];
	}
	++ s[c[u]];
}

int main() {
	scanf("%d", &n);
	rep(i, n) scanf("%d", &c[i]), -- c[i];
	rep(i, n - 1) { int u, v; scanf("%d %d", &u, &v); -- u; -- v; add_edge(u, v); add_edge(v, u); }
	rep(i, n) ans[i] = 1LL * n * (n + 1) / 2;
	dfs(0, -1);
	rep(i, n) ans[i] -= 1LL * (n - s[i]) * (n - s[i] + 1) / 2;
	rep(i, n) printf("%lld\n", ans[i]);
	return 0;
}

### Prompt

Generate 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>
using namespace std;

#define ll long long
#define mp make_pair
#define pb push_back

#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#define rep(i, n) for (int i = 0; i < (int)(n); ++ i)

const int mxn = 2e5 + 5;
int n, c[mxn], hd[mxn], nex[mxn << 1], to[mxn << 1], tot, s[mxn], sz[mxn];
ll ans[mxn];

void add_edge(int u, int v) { nex[++ tot] = hd[u]; hd[u] = tot; to[tot] = v; }

void dfs(int u, int p) {
	sz[u] = 1;
	for (int e = hd[u]; e; e = nex[e]) {
		int v = to[e];
		if (v == p) continue;
		int old = s[c[u]];
		dfs(v, u);
		int nw = s[c[u]];
		int f = sz[v] - (nw - old);
		ans[c[u]] -= 1LL * f * (f + 1) / 2;
		s[c[u]] += f;
		sz[u] += sz[v];
	}
	++ s[c[u]];
}

int main() {
	scanf("%d", &n);
	rep(i, n) scanf("%d", &c[i]), -- c[i];
	rep(i, n - 1) { int u, v; scanf("%d %d", &u, &v); -- u; -- v; add_edge(u, v); add_edge(v, u); }
	rep(i, n) ans[i] = 1LL * n * (n + 1) / 2;
	dfs(0, -1);
	rep(i, n) ans[i] -= 1LL * (n - s[i]) * (n - s[i] + 1) / 2;
	rep(i, n) printf("%lld\n", ans[i]);
	return 0;
}

```