5
27468 7465 74275 40573 40155
112071 76270 244461 264202 132397
1 777133331
2 107454154
3 652330694
4 792720519
 5
78188 56310 79021 70050 65217
115040 5149 128449 98357 36580
1 451393770
2 574046602
3 590130784
4 112514248
 7
1 1 1 1 1 1 1
1 3000000000 3000000000 3000000000 1000000000 1000000000 1000000000
1 1000000000
1 1000000000
1 1000000000
1 1000000000
1 1000000000
1 1000000000
 10
2 8 6 1 2 7 6 9 2 8
4 9 4 3 5 2 9 3 7 3
1 8
2 8
3 8
4 10
5 1
6 4
7 3
8 10
9 2
 11
1 1 1 1 1 1 1 1 1 1 1
1 1000000001 1000000001 1000000001 1000000001 1000000001 1000000001 1000000001 1000000001 1000000001 1000000001
1 1000000000
1 1000000000
1 1000000000
1 1000000000
1 1000000000
1 1000000000
1 1000000000
1 1000000000
1 1000000000
1 1000000000
 5
2 1 1 2 3
1 2 2 2 1
1 2
1 3
2 4
1 4

NO
 NO
 NO
 YES
 NO
 NO

#include <bits/stdc++.h>
using namespace std;
inline long long mod(long long n, long long m) {
  long long ret = n % m;
  if (ret < 0) ret += m;
  return ret;
}
long long gcd(long long a, long long b) {
  return (b == 0LL ? a : gcd(b, a % b));
}
long long exp(long long a, long long b, long long m) {
  if (b == 0LL) return 1LL;
  if (b == 1LL) return mod(a, m);
  long long k = mod(exp(a, b / 2, m), m);
  if (b & 1LL) {
    return mod(a * mod(k * k, m), m);
  } else
    return mod(k * k, m);
}
const long long inf = long long(1e17) + 7;
const long long N = 100100;
long long a[N], b[N];
long long need[N];
long long x[N], k[N];
vector<pair<long long, long long> > g[N];
bool BAD(long long A, long long B) {
  if (A > inf / B) return 1;
  return 0;
}
bool dfs(long long v, long long p = -1) {
  b[v] -= a[v];
  for (auto X : g[v]) {
    long long to = X.first, w = X.second;
    if (!dfs(to, v)) return 0;
    if (b[to] < 0) {
      if (BAD(b[to], w)) {
        return 0;
      }
      b[v] -= (-b[to] * w);
    } else if (b[to] > 0) {
      b[v] += b[to];
    }
  }
  return b[v] >= 0;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> b[i];
  for (long long i = 1; i <= n; i++) cin >> a[i];
  for (long long i = 2; i <= n; i++) {
    cin >> x[i] >> k[i];
    g[x[i]].push_back(pair<long long, long long>(i, k[i]));
  }
  if (dfs(1))
    cout << "YES"
         << "\n";
  else
    cout << "NO"
         << "\n";
}
