#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int max_n = 200005;
int n, c[max_n];
vector<int> G[max_n];
int cur[max_n], sz[max_n];
ll ans[max_n];
void dfs(int u, int fa) {
    int x = cur[c[u]];
    sz[u] = 1;
    for(auto v : G[u]) {
        if(v == fa) continue;
        int pre = cur[c[u]];
        dfs(v, u);
        sz[u] += sz[v];
        int suf = cur[c[u]];
        suf -= pre;
        int t = sz[v] - suf;
        ans[c[u]] -= 1ll * t * (t + 1) / 2;
    }
    x += sz[u];
    cur[c[u]] = x;
}
int main() {
    scanf("%d", &n);
    for(int i = 1; i<= n; i++) scanf("%d", c + i);
    for(int i = 1; i < n; i++) {
        int u, v; scanf("%d%d", &u, &v);
        G[u].push_back(v), G[v].push_back(u);
    }
    dfs(1, 0);
    for(int i = 1; i <= n; i++) {
        int t = n - cur[i];
        ans[i] -= 1ll * t * (t + 1) / 2;
    }
    for(int i = 1; i <= n; i++) ans[i] += 1ll * n * (n + 1) / 2;
    for(int i = 1; i <= n; i++) printf("%lld\n", ans[i]);
    return 0;
}
### Prompt

Develop a solution in CPP 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;
typedef long long ll;
const int max_n = 200005;
int n, c[max_n];
vector<int> G[max_n];
int cur[max_n], sz[max_n];
ll ans[max_n];
void dfs(int u, int fa) {
    int x = cur[c[u]];
    sz[u] = 1;
    for(auto v : G[u]) {
        if(v == fa) continue;
        int pre = cur[c[u]];
        dfs(v, u);
        sz[u] += sz[v];
        int suf = cur[c[u]];
        suf -= pre;
        int t = sz[v] - suf;
        ans[c[u]] -= 1ll * t * (t + 1) / 2;
    }
    x += sz[u];
    cur[c[u]] = x;
}
int main() {
    scanf("%d", &n);
    for(int i = 1; i<= n; i++) scanf("%d", c + i);
    for(int i = 1; i < n; i++) {
        int u, v; scanf("%d%d", &u, &v);
        G[u].push_back(v), G[v].push_back(u);
    }
    dfs(1, 0);
    for(int i = 1; i <= n; i++) {
        int t = n - cur[i];
        ans[i] -= 1ll * t * (t + 1) / 2;
    }
    for(int i = 1; i <= n; i++) ans[i] += 1ll * n * (n + 1) / 2;
    for(int i = 1; i <= n; i++) printf("%lld\n", ans[i]);
    return 0;
}
```