/// Opus 5.5

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

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n;
    cin >> n;
    vector<vector<int>> g(n);
    for (int i = 0; i < n - 1; i++) {
        int u, v;
        cin >> u >> v;
        --u;
        --v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    vector<int> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];

    // BFS 順（再帰を使わない）
    vector<int> par(n, -1), order;
    order.reserve(n);
    par[0] = 0;
    order.push_back(0);
    for (int i = 0; i < (int)order.size(); i++) {
        int v = order[i];
        for (int c : g[v]) {
            if (par[c] == -1) {
                par[c] = v;
                order.push_back(c);
            }
        }
    }

    const long long NEG = LLONG_MIN / 4;
    // P: 純ブロック, U: 未完成ドミノ, S: 完成ドミノ, O: 値1の頂点
    vector<long long> P(n, NEG), U(n, NEG), S(n, NEG), O(n, NEG);

    for (int i = n - 1; i >= 0; i--) {
        int v = order[i];
        long long sumP = 0, sumU = 0, sumO = 0, best = NEG;
        for (int c : g[v]) {
            if (c == par[v]) continue;
            long long cl = max({P[c], S[c], O[c]});  // 辺 (v,c) を切る
            long long pc = max(S[c], O[c]) + 1;       // c を別ピースとしてつなぐ
            if (a[v] == 1) {
                sumO += max(cl, pc);
                continue;
            }
            long long keepP = (a[c] == a[v]) ? P[c] + 1 : NEG;
            sumP += max(cl, keepP);
            long long joinU = (a[c] == a[v]) ? U[c] + 1 : NEG;
            long long h = max({cl, pc, joinU});
            sumU += h;
            long long sat = NEG;  // c を「ドミノを完成させる子」にする場合
            if (a[c] == 2 - a[v]) sat = max(sat, U[c] + 1);
            if (a[c] == a[v]) sat = max(sat, S[c] + 1);
            if (sat >= 0) best = max(best, sat - h);
        }
        if (a[v] == 1) {
            O[v] = sumO;
        } else {
            P[v] = sumP;
            U[v] = sumU;
            if (best != NEG) S[v] = sumU + best;
        }
    }

    cout << max({P[0], S[0], O[0]}) << '\n';
    return 0;
}