#include <bits/stdc++.h>

using namespace std;

#define sz(x) (int) x.size()
#define all(a) a.begin(), a.end()

const int MAXN = 2 * 105000;

int n;
int color[MAXN];
vector<int> g[MAXN];
int sum[MAXN];
long long rem[MAXN];
int sz[MAXN];

void dfs(int v, int par = -1) {
    int curSum = sum[color[v]];
    sz[v] = 1;

    int prev = sum[color[v]];
    for (int to : g[v]) {
        if (to == par)
            continue;

        dfs(to, v);
        if (color[to] != color[v]) {
            int num = sz[to] - (sum[color[v]] - prev);
            rem[color[v]] += 1ll * num * (num + 1) / 2;
        }

        sz[v] += sz[to];
        prev = sum[color[v]];
    }

    sum[color[v]] = curSum + sz[v];
}

int main() {

    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        scanf("%d", &color[i]);
        color[i]--;
    }

    for (int i = 0; i < n - 1; i++) {
        int from, to;
        scanf("%d %d", &from, &to);
        from--; to--;
        g[from].push_back(to);
        g[to].push_back(from);
    }

    dfs(0);
    for (int i = 0; i < n; i++) {
        if (color[0] == i) continue;

        int num = sz[0] - sum[i];
        rem[i] += 1ll * num * (num + 1) / 2;
    }

    for (int i = 0; i < n; i++) 
        printf("%lld\n", 1ll * n * (n + 1) / 2 - rem[i]);

    return 0;
}

### Prompt

Please formulate 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 sz(x) (int) x.size()
#define all(a) a.begin(), a.end()

const int MAXN = 2 * 105000;

int n;
int color[MAXN];
vector<int> g[MAXN];
int sum[MAXN];
long long rem[MAXN];
int sz[MAXN];

void dfs(int v, int par = -1) {
    int curSum = sum[color[v]];
    sz[v] = 1;

    int prev = sum[color[v]];
    for (int to : g[v]) {
        if (to == par)
            continue;

        dfs(to, v);
        if (color[to] != color[v]) {
            int num = sz[to] - (sum[color[v]] - prev);
            rem[color[v]] += 1ll * num * (num + 1) / 2;
        }

        sz[v] += sz[to];
        prev = sum[color[v]];
    }

    sum[color[v]] = curSum + sz[v];
}

int main() {

    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        scanf("%d", &color[i]);
        color[i]--;
    }

    for (int i = 0; i < n - 1; i++) {
        int from, to;
        scanf("%d %d", &from, &to);
        from--; to--;
        g[from].push_back(to);
        g[to].push_back(from);
    }

    dfs(0);
    for (int i = 0; i < n; i++) {
        if (color[0] == i) continue;

        int num = sz[0] - sum[i];
        rem[i] += 1ll * num * (num + 1) / 2;
    }

    for (int i = 0; i < n; i++) 
        printf("%lld\n", 1ll * n * (n + 1) / 2 - rem[i]);

    return 0;
}

```