#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using db = double;
using str = string;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using pd = pair<db, db>;
using vi = vector<int>;
using vb = vector<bool>;
using vl = vector<ll>;
using vd = vector<db>;
using vs = vector<str>;
using vpi = vector<pi>;
using vpl = vector<pl>;
using vpd = vector<pd>;
#define mp make_pair
#define f first
#define s second
#define sz(x) (int)(x).size()
#define bg(x) begin(x)
#define all(x) bg(x), end(x)
#define sor(x) sort(all(x))
#define ft front()
#define bk back()
#define pb push_back
#define pf push_front
#define lb lower_bound
#define ub upper_bound
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)
#define EACH(a,x) for (auto& a: x)
const int MOD = 1e9 + 7;
const int MXN = 1e5 + 10;
const int MXV = 101;
const ll INF = 1e18;
int N, V[MXN]; vi child[MXN]; bitset<MXV> low[MXN], high[MXN]; // # of possible lowests, highests
bitset<MXV> LOW[MXV], HIGH[MXV]; // LOW[C1][C2] means C2 must be the lowest, with C1 as the highest
void solve(int X) {
low[X][V[X]] = high[X][V[X]] = 1; EACH(C, child[X]) solve(C);
FOR(i, 1, MXV) LOW[i].reset(), HIGH[i].reset();
EACH(C, child[X]) FOR(i, 1, MXV) if (low[C][i]) FOR(j, 1, MXV) if (high[C][j]) {
if (j < V[X]) LOW[j][i] = 1; if (i > V[X]) HIGH[i][j] = 1;
}
ROF(i, 2, V[X] + 1) if (low[X][i]) low[X] |= LOW[i - 1];
FOR(i, V[X], MXV - 1) if (high[X][i]) high[X] |= HIGH[i + 1];
}
int main() {
ios_base::sync_with_stdio(false); cin.tie(0);
cin >> N; FOR(i, 1, N + 1) cin >> V[i];
F0R(i, N - 1) { int A, B; cin >> A >> B; child[A].pb(B); }
solve(1); cout << low[1].count() * high[1].count();
}
