Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3
1 2 1
1 2
2 3

Sample Output 1

5
4
0

Let P_{i,j} denote the simple path connecting Vertex i and j.

There are 5 simple paths that visit a vertex painted in the color 1 one or more times:

P_{1,1}\,,\,
P_{1,2}\,,\,
P_{1,3}\,,\,
P_{2,3}\,,\,
P_{3,3}

There are 4 simple paths that visit a vertex painted in the color 2 one or more times:

P_{1,2}\,,\,
P_{1,3}\,,\,
P_{2,2}\,,\,
P_{2,3}

There are no simple paths that visit a vertex painted in the color 3 one or more times.

Sample Input 2

1
1

Sample Output 2

1

Sample Input 3

2
1 2
1 2

Sample Output 3

2
2

Sample Input 4

5
1 2 3 4 5
1 2
2 3
3 4
3 5

Sample Output 4

5
8
10
5
5

Sample Input 5

8
2 7 2 5 4 1 7 5
3 1
1 2
2 7
4 5
5 6
6 8
7 8

Sample Output 5

18
15
0
14
23
0
23
0

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

using i64 = int64_t;

#define rep(i, x, y) for (i64 i = i64(x), i##_max_for_repmacro = i64(y); i < i##_max_for_repmacro; ++i)
#define debug(x) #x << "=" << (x)

#ifdef DEBUG
#define _GLIBCXX_DEBUG
#define print(x) std::cerr << debug(x) << " (L:" << __LINE__ << ")" << std::endl
#else
#define print(x)
#endif

template <i64 p>
class fp {
    public:
    i64 x;
    fp() : x(0) {}
    fp(i64 x_) : x((x_ % p + p) % p) {}
    fp operator+() const { return fp(x); }
    fp operator-() const { return fp(-x); }
    fp& operator+=(const fp& y) {
        x += y.x;
        if (x >= p) x -= p;
        return *this;
    }
    fp& operator-=(const fp& y) { return *this += -y; }
    fp& operator*=(const fp& y) {
        x = x * y.x % p;
        return *this;
    }
    fp& operator/=(const fp& y) { return *this *= fp(inverse(y.x)); }
    fp operator+(const fp& y) const { return fp(x) += y; }
    fp operator-(const fp& y) const { return fp(x) -= y; }
    fp operator*(const fp& y) const { return fp(x) *= y; }
    fp operator/(const fp& y) const { return fp(x) /= y; }
    bool operator==(const fp& y) const { return x == y.x; }
    bool operator!=(const fp& y) const { return !(*this == y); }
    i64 extgcd(i64 a, i64 b, i64& x, i64& y) {
        i64 d = a;
        if (b != 0) {
            d = extgcd(b, a % b, y, x);
            y -= (a / b) * x;
        } else {
            x = 1;
            y = 0;
        }
        return d;
    }
    i64 inverse(i64 a) {
        i64 x, y;
        extgcd(a, p, x, y);
        return (x % p + p) % p;
    }
};

template <i64 p>
i64 abs(const fp<p>& x) { return x.x; }

template <i64 p>
istream& operator>>(istream& is, fp<p>& x) {
    is >> x.x;
    return is;
}

template <i64 p>
ostream& operator<<(ostream& os, const fp<p>& x) {
    os << x.x;
    return os;
}

template <typename T, typename U>
ostream& operator<<(ostream& os, const pair<T, U>& p) {
    os << "(" << p.first << ", " << p.second << ")";
    return os;
}

template <typename T>
ostream& operator<<(ostream& os, const vector<T>& vec) {
    os << "[";
    for (const auto& v : vec) {
        os << v << ",";
    }
    os << "]";
    return os;
}

template <typename T>
bool chmin(T& a, const T& b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}

template <typename T>
bool chmax(T& a, const T& b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}

template <typename A, typename T, size_t size>
void fill(A (&ary)[size], const T& val) {
    fill((T*)ary, (T*)(ary + size), val);
}

constexpr int inf = 1.01e9;
constexpr i64 inf64 = 4.01e18;
constexpr long double eps = 1e-9;

// double(64bit浮動小数)のn分探索のループ回数の上限(2分探索なら50でも十分かもしれない). long double(80ビットの x87 浮動小数点型?)だと, 2分探索であってもこれだと足りないケースがある気がするので, もうちょっと余裕を持たせた方が良さそう.
constexpr i64 max_loop = 100;

template <class T, T op(T, T)>
class segtree {
public:
    int n, size_;
    vector<T> dat;
    T id_;
    segtree() = default;
    segtree(int size, T id, T initial_value) { init(size, id, initial_value); }
    void init(int size, T id, T initial_value) {
        size_ = size;
        id_ = id;
        n = 1;
        while (n < size) n *= 2;
        dat.assign(2 * n - 1, id);
        for (int i = 0; i < size; ++i) update(i, initial_value);
    }
    int size() const { return size_; }
    void update(int k, T a) {
        k += n - 1;  // leaf
        dat[k] = a;
        while (k > 0) {
            k = (k - 1) / 2;
            dat[k] = op(dat[k * 2 + 1], dat[k * 2 + 2]);
        }
    }
    T at(int index) { return dat[index + n - 1]; }
    void add(int k, T a) { update(k, at(k) + a); }
    T query(int a, int b) { return query(a, b, 0, 0, n); }
    T query(int a, int b, int k, int l, int r) {
        if (r <= a or b <= l) return id_;
        if (a <= l and r <= b) return dat[k];
        int m = (l + r) / 2;
        return op(query(a, b, k * 2 + 1, l, m), query(a, b, k * 2 + 2, m, r));
    }
};

template<typename T, T op(T, T)>
struct euler_tour_technique{
    const i64 size; // 頂点数
    vector<vector<i64>> tree;
    const T identity;
    segtree<T,op> seg;
    vector<i64> left,right;
    bool has_inited=false;

    euler_tour_technique(i64 size, const T& identity): size(size), tree(size), identity(identity), seg(2 * size, identity, identity), left(size), right(size){}

    void add_edge(i64 u,i64 v){
        tree[u].emplace_back(v);
        tree[v].emplace_back(u);
    }

    void init(i64 root){
        has_inited=true;

        i64 time=0;
        function<void(i64,i64)> dfs=[&](i64 u,i64 p){
            left[u]=time;
            ++time;
            for(i64 v:tree[u]){
                if(v==p) continue;
                dfs(v,u);
            }
            right[u]=time;
            ++time;
        };
        dfs(root,-1);
    }

    void update(i64 u,T val){
        assert(has_inited);
        seg.update(left[u],val);
    }

    // 注意: 未検証!!!!!!!!!!!!!!!!!!!!!!!
    T at(i64 u){
        assert(has_inited);
        return seg.at(u);
    }

    T subtree_query(i64 u){
        assert(has_inited);
        return seg.query(left[u],right[u]);
    }
};

i64 add(i64 a,i64 b){
    return a+b;
}

void solve() {
    //constexpr i64 mod = 1'000'000'007;

    i64 N;
    cin >> N;

    vector<i64> c(N);
    rep(i,0,N){
        cin >> c[i];
        --c[i];
    }

    vector<vector<i64>> tree(N);
    vector<i64> a(N),b(N);
    rep(i,0,N-1){
        cin >> a[i] >> b[i];
        --a[i];
        --b[i];
        tree[a[i]].emplace_back(b[i]);
        tree[b[i]].emplace_back(a[i]);
    }

    vector<vector<i64>> same_color(N);
    rep(i,0,N) same_color[c[i]].emplace_back(i);

    vector<i64> ans(N,N*(N-1)/2+N);

    vector<i64> depth(N),size(N),parent(N);
    function<void(i64,i64)> calc_depth_size_parent=[&](i64 u, i64 p){
        parent[u]=p;
        size[u]=1;
        for(i64 v:tree[u]){
            if(v==p) continue;
            depth[v]=depth[u]+1;
            calc_depth_size_parent(v, u);
            size[u]+=size[v];
        }
    };
    calc_depth_size_parent(0, -1);

    euler_tour_technique<i64,add> ett(N, 0);
    rep(i,0,N-1) ett.add_edge(a[i],b[i]);
    ett.init(0);
    rep(i,0,N) ett.update(i,1);

    rep(i,0,N){
        vector<i64> vs=same_color[i];
        sort(begin(vs),end(vs),[&](i64 u,i64 v){ return depth[u]>depth[v]; });

        for(i64 u:vs){
            i64 cancellation=0;

            for(i64 v:tree[u]){
                if(v==parent[u]) continue;

                i64 num=ett.subtree_query(v);
                ans[i]-=num*(num+1)/2;

                cancellation+=num;
            }

            ett.update(u,-cancellation);
        }

        if(find(begin(vs),end(vs),0)==end(vs)){
            i64 num=ett.subtree_query(0);
            ans[i]-=num*(num+1)/2;
        }

        for(i64 u:vs){
            ett.update(u,1);
        }
    }

    rep(i,0,N) cout << ans[i] << endl;
}

int main() {
    std::cin.tie(0);
    std::ios::sync_with_stdio(false);
    cout.setf(ios::fixed);
    cout.precision(16);
    solve();
    return 0;
}