#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 = 5005 + 9, mod = 998244353;
const ll OO = 1e9;

int n, v[N];
bool factors[10005][10005], prime[10005];
ll dp[N][10005];
vector<int> g[N];

ll solve(int u, int p, int last) {
    auto &ret = dp[u][last];
    if (~ret) {
        return ret;
    }
    ret = OO;
    for (int i = 1; i <= 10000; ++i) {
        auto x = i;
        auto y = last;
        if (x > y) {
            swap(x, y);
        }
        if (last != 0 && !factors[x][y]) {
            continue;
        }
        ll cur = 0;
        if (i != v[u]) {
            cur += i;
        }
        for (auto &nxt: g[u]) {
            if (nxt != p) {
                cur += solve(nxt, u, i);
            }
        }
        ret = min(ret, cur);
    }
    return ret;
}

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    fill(prime + 1, prime + 10001, 1);
    for (int i = 2; i <= 10000; ++i) {
        if (prime[i]) {
            vector<int> nums;
            for (int j = i; j <= 10000; j += i) {
                if (j != i) {
                    prime[j] = false;
                }
                nums.push_back(j);
            }
            for (int j = 0; j < nums.size(); ++j) {
                for (int k = j; k < nums.size(); ++k) {
                    factors[nums[j]][nums[k]] = true;
                }
            }
        }
    }
    memset(dp, -1, sizeof(dp));
    cin >> n;
    for (int i = 1; i <= n; ++i) {
        cin >> v[i];
    }
    for (int i = 0; i < n - 1; ++i) {
        int a, b;
        cin >> a >> b;
        g[a].push_back(b);
        g[b].push_back(a);
    }
    cout << solve(1, 0, 0) << '\n';
    return 0;
}