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;
const int N = 1e5 + 5;
const long long inf = 1e18;
long long materials[N];
vector<vector<pair<int, long long>>> edge(N);
int n;
pair<long long, long long> dfs(int now) {
  long long need = max(-materials[now], 0LL);
  long long supply = max(materials[now], 0LL);
  for (pair<int, long long> it : edge[now]) {
    pair<long long, long long> ret = dfs(it.first);
    if (ret.first > inf / it.second) {
      cout << "NO" << endl;
      exit(0);
    } else {
      need = need + ret.first * it.second;
    }
    if (need > inf) {
      cout << "NO" << endl;
      exit(0);
    }
    supply = supply + ret.second;
  }
  if (need < supply) {
    return {0, supply - need};
  } else if (supply < need) {
    return {need - supply, 0};
  } else {
    return {0, 0};
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    cin >> materials[i];
  }
  for (int i = 1; i <= n; i++) {
    long long temp;
    cin >> temp;
    materials[i] = materials[i] - temp;
  }
  for (int i = 2; i <= n; i++) {
    int x, k;
    scanf("%d%d", &x, &k);
    edge[x].emplace_back(i, k);
  }
  pair<long long, long long> ret = dfs(1);
  if (ret.first > 0)
    cout << "NO" << endl;
  else
    cout << "YES" << endl;
}
