#include <bits/stdc++.h>
#define rep(i,a,b) for(int i = a; i < b; ++i)
#define all(c) c.begin(), c.end()
#define gmax(x,y) x=max(x,y)
#define gmin(x,y) x=min(x,y)
#define gadd(x,y) x=add(x,y)
#define gmul(x,y) x=mul(x,y)
using namespace std;

typedef pair<int,int> pii;
typedef long long ll;
typedef vector<int> vi;

const int N = 2e5 + 5;

vector<int> g[N], h[2*N];
int c[N], cind[N], sz[N], par[N];

void init(int u, int p=0){
	int cpar = cind[c[u]];	
	h[cpar].push_back(u);
	sz[u] = 1;
	par[u] = p;
	for(int v:g[u]){
		if(v == p)continue;
		cind[c[u]] = v;
		init(v,u);
		sz[u] += sz[v];
	}
	cind[c[u]] = cpar;
}

ll ans[N];

int main(){
	cin.tie(0);
	cout.tie(0);
	ios_base::sync_with_stdio(0);
	int n;
	cin >> n;
	rep(i,1,n+1){
		cin >> c[i];
		c[i]--;
	}
	rep(i,0,n-1){
		int x,y;
		cin >> x >> y;
		g[x].push_back(y);
		g[y].push_back(x);
	}
	iota(cind,cind+N,N);
	init(1);
	rep(i,2,n+1){
		int ctar = c[par[i]];	
		ll cnt = sz[i];
		for(int v:h[i]){
			cnt -= sz[v];
		}
		ans[ctar] += cnt*(cnt-1)/2 + cnt;
	}
	rep(i,N,2*N){
		int ctar = i - N;
		ll cnt = sz[1];
		for(int v:h[i]){
			cnt -= sz[v];
		}
		ans[ctar] += cnt*(cnt-1)/2 + cnt;
	}
	rep(i,0,n)ans[i] = 1LL * n * (n-1)/2 + n - ans[i];
	rep(i,0,n)cout << ans[i] << '\n';
}

### Prompt

Your challenge is to write 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>
#define rep(i,a,b) for(int i = a; i < b; ++i)
#define all(c) c.begin(), c.end()
#define gmax(x,y) x=max(x,y)
#define gmin(x,y) x=min(x,y)
#define gadd(x,y) x=add(x,y)
#define gmul(x,y) x=mul(x,y)
using namespace std;

typedef pair<int,int> pii;
typedef long long ll;
typedef vector<int> vi;

const int N = 2e5 + 5;

vector<int> g[N], h[2*N];
int c[N], cind[N], sz[N], par[N];

void init(int u, int p=0){
	int cpar = cind[c[u]];	
	h[cpar].push_back(u);
	sz[u] = 1;
	par[u] = p;
	for(int v:g[u]){
		if(v == p)continue;
		cind[c[u]] = v;
		init(v,u);
		sz[u] += sz[v];
	}
	cind[c[u]] = cpar;
}

ll ans[N];

int main(){
	cin.tie(0);
	cout.tie(0);
	ios_base::sync_with_stdio(0);
	int n;
	cin >> n;
	rep(i,1,n+1){
		cin >> c[i];
		c[i]--;
	}
	rep(i,0,n-1){
		int x,y;
		cin >> x >> y;
		g[x].push_back(y);
		g[y].push_back(x);
	}
	iota(cind,cind+N,N);
	init(1);
	rep(i,2,n+1){
		int ctar = c[par[i]];	
		ll cnt = sz[i];
		for(int v:h[i]){
			cnt -= sz[v];
		}
		ans[ctar] += cnt*(cnt-1)/2 + cnt;
	}
	rep(i,N,2*N){
		int ctar = i - N;
		ll cnt = sz[1];
		for(int v:h[i]){
			cnt -= sz[v];
		}
		ans[ctar] += cnt*(cnt-1)/2 + cnt;
	}
	rep(i,0,n)ans[i] = 1LL * n * (n-1)/2 + n - ans[i];
	rep(i,0,n)cout << ans[i] << '\n';
}

```