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

const int maxn = 1e6 + 5;
int n;
vector<int> v[maxn];
int cnt_son[maxn];
#define ll long long 
ll ans[maxn];

ll cal(ll x) {
    return x* (x + 1) / 2;
}

int col[maxn];
int num[maxn];

void dfs(int x, int pre) {
    cnt_son[x] = 1;
    int c = col[x];
    int temp = num[c];
    for (auto &y : v[x]) {
        if (y != pre) {
            int now = num[c];
            dfs(y, x);
            now = num[c] - now;
            ans[c] -= cal(cnt_son[y] - now);
            cnt_son[x] += cnt_son[y];
        }
    }
    num[c] = temp + cnt_son[x];
    return ;
}


int main() {
    cin >> n;
    int x, y;
    for (int i = 1; i <= n; ++i) {
        scanf("%d", &col[i]);
    }
    for (int i = 1; i < n; ++i) {
        scanf("%d %d", &x, &y);
        v[x].push_back(y);
        v[y].push_back(x);
    }
    for (int i = 1; i <= n; ++i) {
        ans[i] = cal(n);
    }
    dfs(1, 0);
    for (int i = 1; i <= n; ++i) {
        ans[i] -= cal(n -num[i]);
        printf("%lld\n", ans[i]);
    }
}
### 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;

const int maxn = 1e6 + 5;
int n;
vector<int> v[maxn];
int cnt_son[maxn];
#define ll long long 
ll ans[maxn];

ll cal(ll x) {
    return x* (x + 1) / 2;
}

int col[maxn];
int num[maxn];

void dfs(int x, int pre) {
    cnt_son[x] = 1;
    int c = col[x];
    int temp = num[c];
    for (auto &y : v[x]) {
        if (y != pre) {
            int now = num[c];
            dfs(y, x);
            now = num[c] - now;
            ans[c] -= cal(cnt_son[y] - now);
            cnt_son[x] += cnt_son[y];
        }
    }
    num[c] = temp + cnt_son[x];
    return ;
}


int main() {
    cin >> n;
    int x, y;
    for (int i = 1; i <= n; ++i) {
        scanf("%d", &col[i]);
    }
    for (int i = 1; i < n; ++i) {
        scanf("%d %d", &x, &y);
        v[x].push_back(y);
        v[y].push_back(x);
    }
    for (int i = 1; i <= n; ++i) {
        ans[i] = cal(n);
    }
    dfs(1, 0);
    for (int i = 1; i <= n; ++i) {
        ans[i] -= cal(n -num[i]);
        printf("%lld\n", ans[i]);
    }
}
```