#include <stdio.h>
#include <vector>
using namespace std;
vector<int> e[200010];
int c[200010];
long long cnt[200010];
long long res[200010];
long long s[200010];
void dfs(int fa, int x) {
    int oriCnt = cnt[c[x]];
    s[x] = 1;
    for (int y : e[x]) {
        if (y == fa) continue;
        int preCnt = cnt[c[x]];
        dfs(x, y);
        long long num = s[y] - (cnt[c[x]] - preCnt);
        res[c[x]] += (num * (num-1))/2+num;
        s[x] += s[y];
    }
    cnt[c[x]] = oriCnt + s[x];
}
int main() {
    int n, i, x, y;
    scanf("%d", &n);
    for (i = 1; i <= n; i++) {
        scanf("%d", &c[i]);
    }
    for (i = 1; i < n; i++) {
        scanf("%d%d", &x, &y);
        e[x].push_back(y);
        e[y].push_back(x);
    }
    dfs(0, 1);
    for (i = 1; i <= n; i++) {
        long long num = s[1] - cnt[i];
        res[i] += (num * (num-1))/2+num;
        printf("%lld\n", (1ll*n*(n-1))/2+n-res[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 <stdio.h>
#include <vector>
using namespace std;
vector<int> e[200010];
int c[200010];
long long cnt[200010];
long long res[200010];
long long s[200010];
void dfs(int fa, int x) {
    int oriCnt = cnt[c[x]];
    s[x] = 1;
    for (int y : e[x]) {
        if (y == fa) continue;
        int preCnt = cnt[c[x]];
        dfs(x, y);
        long long num = s[y] - (cnt[c[x]] - preCnt);
        res[c[x]] += (num * (num-1))/2+num;
        s[x] += s[y];
    }
    cnt[c[x]] = oriCnt + s[x];
}
int main() {
    int n, i, x, y;
    scanf("%d", &n);
    for (i = 1; i <= n; i++) {
        scanf("%d", &c[i]);
    }
    for (i = 1; i < n; i++) {
        scanf("%d%d", &x, &y);
        e[x].push_back(y);
        e[y].push_back(x);
    }
    dfs(0, 1);
    for (i = 1; i <= n; i++) {
        long long num = s[1] - cnt[i];
        res[i] += (num * (num-1))/2+num;
        printf("%lld\n", (1ll*n*(n-1))/2+n-res[i]);
    }
    return 0;
}

```