Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3
1 2 1
1 2
2 3

Sample Output 1

5
4
0

Let P_{i,j} denote the simple path connecting Vertex i and j.

There are 5 simple paths that visit a vertex painted in the color 1 one or more times:

P_{1,1}\,,\,
P_{1,2}\,,\,
P_{1,3}\,,\,
P_{2,3}\,,\,
P_{3,3}

There are 4 simple paths that visit a vertex painted in the color 2 one or more times:

P_{1,2}\,,\,
P_{1,3}\,,\,
P_{2,2}\,,\,
P_{2,3}

There are no simple paths that visit a vertex painted in the color 3 one or more times.

Sample Input 2

1
1

Sample Output 2

1

Sample Input 3

2
1 2
1 2

Sample Output 3

2
2

Sample Input 4

5
1 2 3 4 5
1 2
2 3
3 4
3 5

Sample Output 4

5
8
10
5
5

Sample Input 5

8
2 7 2 5 4 1 7 5
3 1
1 2
2 7
4 5
5 6
6 8
7 8

Sample Output 5

18
15
0
14
23
0
23
0

Solution in C (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <string.h>

int n;
int a[200001];
int b[200001];
int c[200001];
int f[200001];

int dfs( int nd )
{
    int rv = 0;
    f[nd] = 1;

    for( int j=0; j<n-1; j++ ) {
        if( a[j] == nd && f[b[j]] == 0 ) rv += dfs( b[j] );
        if( b[j] == nd && f[a[j]] == 0 ) rv += dfs( a[j] );
    }

    return( rv + 1 );
}

int main( int argc, char *argv[] )
{
    int m;
    int t;

    scanf( "%d", &n );
    for( int i=1; i<=n; i++ )
        scanf( "%d", &c[i] );

    for( int i=0; i<n-1; i++ )
        scanf( "%d %d", &a[i], &b[i] );

    for( int i=1; i<=n; i++ ) {

        t = n * (n+1) / 2;

        for( int j=1; j<=n; j++ ) {
            if( c[j] == i ) f[j] = 1;
            else            f[j] = 0;
        }

        for( int j=1; j<=n; j++ ) {
            if( f[j] == 0 ) {
                m = dfs( j );
                t -= m * (m + 1) / 2;
            }
        }

        printf( "%d\n", t );
    }
}