#include <bits/stdc++.h>
using namespace std;
#define rep2(i, m, n) for(int i=int(m); i<int(n); ++i)
#define rep(i, n) rep2(i, 0, n)
using ll = long long;
using V = vector<int>;
using Vll = vector<ll>;
using VV = vector<V>;
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;

VV G;
V c, dp;
Vll ans;
int s = 0;

ll chs(ll x) {
  return x*(x+1)/2;
}

void dfs(int u, int p=-1) {
  int tmp = s + dp[c[u]];
  ++s;
  for (auto v : G[u]) if (v != p) {
    dp[c[u]] = -s;
    dfs(v, u);
    ans[c[u]] -= chs(s + dp[c[u]]);
  }
  dp[c[u]] = tmp - s;
}

int main() {
  ll n; cin >> n;
  G.assign(n, V());
  c.assign(n, 0);
  dp.assign(n, 0);
  ans.assign(n, n*(n+1)/2);

  rep(i, n) {
    cin >> c[i];
    --c[i];
  }
  rep(i, n-1) {
    int a, b; cin >> a >> b;
    a--; b--;
    G[a].push_back(b);
    G[b].push_back(a);
  }

  dfs(0);
  rep(i, n) {
    ans[i] -= chs(s + dp[i]);
    cout << ans[i] << endl;
  }
  return 0;
}

### Prompt

Develop a solution in cpp 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 rep2(i, m, n) for(int i=int(m); i<int(n); ++i)
#define rep(i, n) rep2(i, 0, n)
using ll = long long;
using V = vector<int>;
using Vll = vector<ll>;
using VV = vector<V>;
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;

VV G;
V c, dp;
Vll ans;
int s = 0;

ll chs(ll x) {
  return x*(x+1)/2;
}

void dfs(int u, int p=-1) {
  int tmp = s + dp[c[u]];
  ++s;
  for (auto v : G[u]) if (v != p) {
    dp[c[u]] = -s;
    dfs(v, u);
    ans[c[u]] -= chs(s + dp[c[u]]);
  }
  dp[c[u]] = tmp - s;
}

int main() {
  ll n; cin >> n;
  G.assign(n, V());
  c.assign(n, 0);
  dp.assign(n, 0);
  ans.assign(n, n*(n+1)/2);

  rep(i, n) {
    cin >> c[i];
    --c[i];
  }
  rep(i, n-1) {
    int a, b; cin >> a >> b;
    a--; b--;
    G[a].push_back(b);
    G[b].push_back(a);
  }

  dfs(0);
  rep(i, n) {
    ans[i] -= chs(s + dp[i]);
    cout << ans[i] << endl;
  }
  return 0;
}

```