#pragma GCC optimize("O3")
#pragma GCC optimize ("unroll-loops")
#pragma GCC target("avx,avx2,fma")

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

#define ll long long
#define ld long double


const int N = 10000 + 5, mod = 998244353;

vector<int> factors[N];
bool vis[N];

bool isPrime(int n) {
    for(int i = 2; i * i <= n; i++) {
        if(n % i == 0) {
            return 0;
        }
    }
    return 1;
}

vector<int> all, primes;


void solve(int idx, int val) {
    if(val > 10000) return;
    if(idx == (int) primes.size()) {
        if(val > 1) all.push_back(val);
        return;
    }

    solve(idx + 1, val);
    solve(idx, val * primes[idx]);
}

int n, val[N], dp[5005][N];
vector<int> adj[N];
vector<int> childSorted[N];

void dfs(int node, int par) {
    dp[node][1] = 2 * n;

    for (int i = 2; i <= 2 * n; i++) {
        if (val[node] % i > 0) dp[node][i] = i;
    }

    for (auto i: adj[node]) {
        if (i == par) continue;
        dfs(i, node);
        memset(vis, 0, sizeof vis);
        for (int j = 0; j <= 2 * n; j++) {
            childSorted[j].clear();
        }
        for (int j = 0; j <= 2 * n; j++) {
            childSorted[dp[i][j]].push_back(j);
        }

        for (int j = 0; j <= 2 * n; j++) {
            for (auto childVal: childSorted[j]) {
                for (auto prime: factors[childVal]) {
                    if (vis[prime]) continue;
                    vis[prime] = 1;
                    for (int k = prime; k <= 2 * n; k += prime) {
                        dp[node][k] += j;
                    }
                }
            }
        }
    }

    for(int i = 1; i <= n; i++) {
        dp[node][i] = min(dp[node][i], 2 * n);
    }
}

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);

    for(int i = 2; i < 100; i++) {
        if(isPrime(i)) {
            primes.push_back(i);
        }
    }

    // solve(0, 1);
    // cout << (int) all.size();

    for(auto i: primes) {
        for(int j = i; j < N; j += i) {
            factors[j].push_back(i);
        }
    }

    cin >> n;
    for(int i = 1; i <= n; i++) cin >> val[i];

    for(int i = 1; i < n; i++) {
        int u, v;
        cin >> u >> v;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }

    dfs(1, 0);

    int ans = 2 * n;
    for(int i = 1; i <= 2 * n; i++) ans = min(ans, dp[1][i]);
    cout << ans;





    return 0;
}