#include<bits/stdc++.h>
using namespace std;
#define LL long long int
#define MAXN 200000
#define DEBUG true

/*
sz[u] -> Size of subtree rooted at node u
cnt[c] -> Number of nodes below a subtree rooted at node with color c
sum[c] -> Number of paths which doesn't go through a node with color c
*/

int N, c[MAXN + 3], cnt[MAXN + 3], sz[MAXN + 3];
LL sum[MAXN + 3];
vector<int> adj[MAXN + 3];

LL C2(int n) {
    return ((LL) n * n + n)/2;
}

void DFS(int u, int prv) {
    sz[u] = 1;
    for (auto v: adj[u]) {
        if (v == prv) continue;

        int prvCnt = cnt[c[u]];

        DFS(v, u);

        int cntOther = sz[v] - (cnt[c[u]] - prvCnt);

        sum[c[u]] += C2(cntOther);
        cnt[c[u]] += cntOther;

        sz[u] += sz[v];
    }

    cnt[c[u]] += 1;
}

int main() {
    cin >> N;
    for (int i = 1; i <= N; i++)
        cin >> c[i];

    for (int i = 1; i < N; i++) {
        int u, v;
        cin >> u >> v;

        adj[u].push_back(v);
        adj[v].push_back(u);
    }

    memset(sz, 0, sizeof(sz));
    memset(cnt, 0, sizeof(cnt));
    memset(sum, 0, sizeof(sum));

    DFS(1, 1);

    for (int c = 1; c <= N; c++) {
        LL all = C2(N);
        LL temp = (sz[1] - cnt[c]);
        cout << all - (sum[c] + C2(temp)) << endl;
    }
}

### 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 LL long long int
#define MAXN 200000
#define DEBUG true

/*
sz[u] -> Size of subtree rooted at node u
cnt[c] -> Number of nodes below a subtree rooted at node with color c
sum[c] -> Number of paths which doesn't go through a node with color c
*/

int N, c[MAXN + 3], cnt[MAXN + 3], sz[MAXN + 3];
LL sum[MAXN + 3];
vector<int> adj[MAXN + 3];

LL C2(int n) {
    return ((LL) n * n + n)/2;
}

void DFS(int u, int prv) {
    sz[u] = 1;
    for (auto v: adj[u]) {
        if (v == prv) continue;

        int prvCnt = cnt[c[u]];

        DFS(v, u);

        int cntOther = sz[v] - (cnt[c[u]] - prvCnt);

        sum[c[u]] += C2(cntOther);
        cnt[c[u]] += cntOther;

        sz[u] += sz[v];
    }

    cnt[c[u]] += 1;
}

int main() {
    cin >> N;
    for (int i = 1; i <= N; i++)
        cin >> c[i];

    for (int i = 1; i < N; i++) {
        int u, v;
        cin >> u >> v;

        adj[u].push_back(v);
        adj[v].push_back(u);
    }

    memset(sz, 0, sizeof(sz));
    memset(cnt, 0, sizeof(cnt));
    memset(sum, 0, sizeof(sum));

    DFS(1, 1);

    for (int c = 1; c <= N; c++) {
        LL all = C2(N);
        LL temp = (sz[1] - cnt[c]);
        cout << all - (sum[c] + C2(temp)) << endl;
    }
}

```