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;
using LL = long long;
constexpr int N = 1e5 + 5;
constexpr LL INF = 2e17;
LL have[N], need[N];
int x[N], k[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> have[i];
  for (int i = 1; i <= n; i++) cin >> need[i];
  for (int i = 2; i <= n; i++) cin >> x[i] >> k[i];
  for (int i = n; i > 1; i--) {
    if (have[i] < need[i]) {
      LL v = need[i] - have[i];
      if ((__int128)v * k[i] > INF) {
        cout << "NO\n";
        return 0;
      }
      have[x[i]] -= (need[i] - have[i]) * k[i];
      have[i] = need[i];
    }
    if (have[i] > need[i]) {
      have[x[i]] += have[i] - need[i];
    }
    if (have[x[i]] < -INF) {
      cout << "NO\n";
      return 0;
    }
  }
  if (have[1] < need[1])
    cout << "NO\n";
  else
    cout << "YES\n";
}
