#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 2e5 + 5;
vector<int> G[N];
int C[N], num[N], sum[N];
ll ans[N];
ll cal(int x) {
    return 1LL * x * (x + 1) / 2;
}
void dfs(int u, int fa) {
    int c = C[u], save = sum[c];
    num[u] = 1;
    for(auto v : G[u]) {
        if(v == fa) continue;
        int t = sum[c];
        dfs(v, u);
        int dt = sum[c] - t;
        ans[c] -= cal(num[v] - dt);
        num[u] += num[v];
    }
    sum[c] = save + num[u];
}
int main() {
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) {
        scanf("%d", &C[i]);
    }
    for(int i = 1, u, v; i < n; i++) {
        scanf("%d%d", &u, &v);
        G[u].push_back(v);
        G[v].push_back(u);
    }
    for(int i = 1; i <= n; i++) {
        ans[i] = cal(n);
    }
    dfs(1, -1);
    for(int i = 1; i <= n; i++) {
        int t = n - sum[i];
        ans[i] -= cal(t);
        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>
#define ll long long
using namespace std;
const int N = 2e5 + 5;
vector<int> G[N];
int C[N], num[N], sum[N];
ll ans[N];
ll cal(int x) {
    return 1LL * x * (x + 1) / 2;
}
void dfs(int u, int fa) {
    int c = C[u], save = sum[c];
    num[u] = 1;
    for(auto v : G[u]) {
        if(v == fa) continue;
        int t = sum[c];
        dfs(v, u);
        int dt = sum[c] - t;
        ans[c] -= cal(num[v] - dt);
        num[u] += num[v];
    }
    sum[c] = save + num[u];
}
int main() {
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) {
        scanf("%d", &C[i]);
    }
    for(int i = 1, u, v; i < n; i++) {
        scanf("%d%d", &u, &v);
        G[u].push_back(v);
        G[v].push_back(u);
    }
    for(int i = 1; i <= n; i++) {
        ans[i] = cal(n);
    }
    dfs(1, -1);
    for(int i = 1; i <= n; i++) {
        int t = n - sum[i];
        ans[i] -= cal(t);
        printf("%lld\n", ans[i]);
    }
    return 0;
}
```