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>
#pragma GCC optimize("O")
using namespace std;
long long a[100005], b[100005], k[100005];
int pa[100005];
int sons[100005];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  for (int i = 1; i <= n; ++i) cin >> b[i];
  for (int i = 2; i <= n; ++i) {
    cin >> pa[i] >> k[i];
    ++sons[pa[i]];
  }
  queue<int> le;
  for (int i = 1; i <= n; ++i)
    if (sons[i] == 0) le.push(i);
  while (le.size()) {
    int now = le.front();
    le.pop();
    if (now == 1) {
      if (a[now] >= b[now])
        cout << "YES" << endl;
      else
        cout << "NO" << endl;
      break;
    }
    if (b[now] == a[now])
      ;
    else if (b[now] > a[now]) {
      long long need = b[now] - a[now];
      long long tk = need * k[now];
      if (tk / need != k[now]) exit((cout << "NO" << endl, 0));
      a[pa[now]] -= tk;
      if (a[pa[now]] < -2e17) exit((cout << "NO" << endl, 0));
    } else {
      long long mo = a[now] - b[now];
      a[pa[now]] += mo;
    }
    --sons[pa[now]];
    if (sons[pa[now]] == 0) le.push(pa[now]);
  }
}
