#include <bits/stdc++.h>
#define FOR(i,k,n)  for(int i = (k);i < (n);++i)
#define REP(i,n)    FOR(i,0,n)
#define ALL(x) begin(x),end(x)
 
using namespace std;
using ll = int64_t;
using vecint = vector<int>;
using vecll = vector<ll>;
 
pair<map<int,ll>, int> dfs(const vector<vecint>& g, const vecint& clr,
    vecll& ans, int i, int p = -1) {
  ll sz = 1;
  map<int,ll> ret;
  for (auto&& to : g[i]) {
    if(to == p) continue;
    auto [cret, csz] = dfs(g, clr, ans, to, i);
    ll icsz = csz - cret[clr[i]];
    if (cret.size() > ret.size()) swap(cret, ret);
    for (auto&& [c, num] : cret) {
      ret[c] += num;
    }
    ans[clr[i]] -= icsz * (icsz + 1) / 2;
    sz += csz;
  }
  ret[clr[i]] = sz;
  return make_pair(move(ret), sz);
}
 
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  ll n;
  cin>>n;
  vecint clr(n);
  REP(i,n)cin>>clr[i];
  vector<vecint> g(n);
  REP(i,n-1) {
    int a,b;
    cin>>a>>b;
    --a;--b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  vecll ans(n+1, n*(n+1)/2);
  auto [cret, _] = dfs(g, clr, ans, 0);
  FOR(i,1,n+1) {
    ll icsz = n - cret[i];
    ans[i] -= icsz * (icsz+1) / 2;
    cout<<ans[i]<<endl;
  }
  return 0;
}
### Prompt

Please create a solution in CPP 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 FOR(i,k,n)  for(int i = (k);i < (n);++i)
#define REP(i,n)    FOR(i,0,n)
#define ALL(x) begin(x),end(x)
 
using namespace std;
using ll = int64_t;
using vecint = vector<int>;
using vecll = vector<ll>;
 
pair<map<int,ll>, int> dfs(const vector<vecint>& g, const vecint& clr,
    vecll& ans, int i, int p = -1) {
  ll sz = 1;
  map<int,ll> ret;
  for (auto&& to : g[i]) {
    if(to == p) continue;
    auto [cret, csz] = dfs(g, clr, ans, to, i);
    ll icsz = csz - cret[clr[i]];
    if (cret.size() > ret.size()) swap(cret, ret);
    for (auto&& [c, num] : cret) {
      ret[c] += num;
    }
    ans[clr[i]] -= icsz * (icsz + 1) / 2;
    sz += csz;
  }
  ret[clr[i]] = sz;
  return make_pair(move(ret), sz);
}
 
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  ll n;
  cin>>n;
  vecint clr(n);
  REP(i,n)cin>>clr[i];
  vector<vecint> g(n);
  REP(i,n-1) {
    int a,b;
    cin>>a>>b;
    --a;--b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  vecll ans(n+1, n*(n+1)/2);
  auto [cret, _] = dfs(g, clr, ans, 0);
  FOR(i,1,n+1) {
    ll icsz = n - cret[i];
    ans[i] -= icsz * (icsz+1) / 2;
    cout<<ans[i]<<endl;
  }
  return 0;
}
```