#pragma GCC optimize("Ofast,inline,unroll-loops,fast-math")
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <vector>
#define N 200020
using namespace std;

int a[N];
vector<int> e[N];
int s[N], c[N];
long long o[N];

inline long long C2(int n)
{
    return (long long)n * (n + 1) / 2;
}

void DFS(int x, int p)
{
    int i, t, r;

    r = c[a[x]];
    s[x] = 1;
    for(i = 0; i < (signed)e[x].size(); i ++)
        if(e[x][i] != p)
        {
            t = c[a[x]];
            DFS(e[x][i], x);
            t = c[a[x]] - t;
            o[a[x]] -= C2(s[e[x][i]] - t);
            s[x] += s[e[x][i]];
        }
    c[a[x]] = r + s[x];

    return;
}

int main(void)
{
    int n, u, v;
    int i;

    scanf("%d", &n);
    for(i = 1; i <= n; i ++)
        scanf("%d", &a[i]);
    for(i = 1; i < n; i ++)
    {
        scanf("%d %d", &u, &v);
        e[u].push_back(v);
        e[v].push_back(u);
    }

    for(i = 1; i <= n; i ++)
        o[i] = C2(n);
    DFS(1, 1);
    for(i = 1; i <= n; i ++)
        printf("%lld\n", o[i] - C2(n - c[i]));

    return 0;
}

### Prompt

Your task is to create 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
#pragma GCC optimize("Ofast,inline,unroll-loops,fast-math")
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <vector>
#define N 200020
using namespace std;

int a[N];
vector<int> e[N];
int s[N], c[N];
long long o[N];

inline long long C2(int n)
{
    return (long long)n * (n + 1) / 2;
}

void DFS(int x, int p)
{
    int i, t, r;

    r = c[a[x]];
    s[x] = 1;
    for(i = 0; i < (signed)e[x].size(); i ++)
        if(e[x][i] != p)
        {
            t = c[a[x]];
            DFS(e[x][i], x);
            t = c[a[x]] - t;
            o[a[x]] -= C2(s[e[x][i]] - t);
            s[x] += s[e[x][i]];
        }
    c[a[x]] = r + s[x];

    return;
}

int main(void)
{
    int n, u, v;
    int i;

    scanf("%d", &n);
    for(i = 1; i <= n; i ++)
        scanf("%d", &a[i]);
    for(i = 1; i < n; i ++)
    {
        scanf("%d %d", &u, &v);
        e[u].push_back(v);
        e[v].push_back(u);
    }

    for(i = 1; i <= n; i ++)
        o[i] = C2(n);
    DFS(1, 1);
    for(i = 1; i <= n; i ++)
        printf("%lld\n", o[i] - C2(n - c[i]));

    return 0;
}

```