#include<bits/stdc++.h>
using namespace std;
 
#define pb push_back
#define ll long long
#define fr(i,j,k) for(int i=j;i<k;i++)
#define f(n) fr(i,0,n)
#define f1(n) fr(i,1,n+1)
#define ms(i) memset(i,0,sizeof(i));
#define ms1(i) memset(i,-1,sizeof(i));
#define pll pair<ll,ll>
#define pii pair<int,int>
#define F first
#define S second
const int maxn = 2e5 + 5;
const int mod = 998244353;
ll ans[200005];
int c[200005];
ll cnt[200005];
ll cur[200005];
ll sz[200005];
vector<int>g[200005];
void dfs(int now, int pre) {
    sz[now] = 1;
    int in = cur[c[now]];
    for (auto i : g[now]) {
        if (i == pre) continue;
        int be = cur[c[now]];
        dfs(i, now);
        int af = cur[c[now]];
        int mt = sz[i] - (af-be);
        ans[c[now]] -= cnt[mt];
        sz[now] += sz[i];
    }
    cur[c[now]] = in + sz[now];
}
int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    ll n;
    cin >> n;
    f1 (n) {
        ans[i] = n * (n + 1) / 2;
        cnt[i] = (ll)i * (i + 1) / 2;
        cin >> c[i];
    }
    f (n - 1) {
        int x, y;
        cin >> x >> y;
        g[x].pb(y);
        g[y].pb(x);
    }
    dfs(1, 0);
    f1 (n) {
        ans[i] -= cnt[n - cur[i]];
        cout << ans[i] << '\n';
    }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 pb push_back
#define ll long long
#define fr(i,j,k) for(int i=j;i<k;i++)
#define f(n) fr(i,0,n)
#define f1(n) fr(i,1,n+1)
#define ms(i) memset(i,0,sizeof(i));
#define ms1(i) memset(i,-1,sizeof(i));
#define pll pair<ll,ll>
#define pii pair<int,int>
#define F first
#define S second
const int maxn = 2e5 + 5;
const int mod = 998244353;
ll ans[200005];
int c[200005];
ll cnt[200005];
ll cur[200005];
ll sz[200005];
vector<int>g[200005];
void dfs(int now, int pre) {
    sz[now] = 1;
    int in = cur[c[now]];
    for (auto i : g[now]) {
        if (i == pre) continue;
        int be = cur[c[now]];
        dfs(i, now);
        int af = cur[c[now]];
        int mt = sz[i] - (af-be);
        ans[c[now]] -= cnt[mt];
        sz[now] += sz[i];
    }
    cur[c[now]] = in + sz[now];
}
int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    ll n;
    cin >> n;
    f1 (n) {
        ans[i] = n * (n + 1) / 2;
        cnt[i] = (ll)i * (i + 1) / 2;
        cin >> c[i];
    }
    f (n - 1) {
        int x, y;
        cin >> x >> y;
        g[x].pb(y);
        g[y].pb(x);
    }
    dfs(1, 0);
    f1 (n) {
        ans[i] -= cnt[n - cur[i]];
        cout << ans[i] << '\n';
    }
}

```