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>
const double Pi = acos(-1.0);
using namespace std;
const int maxN = 100005;
const long long inf = (long long)1e15;
int n;
long long b[maxN];
long long k[maxN];
long long req[maxN];
vector<int> G[maxN];
void dfs(int cur) {
  for (int i = 0; i < (int)G[cur].size(); i++) {
    int nxt = G[cur][i];
    dfs(nxt);
    if (req[nxt] < 0) {
      if (inf / k[nxt] < (-req[nxt])) {
        puts("NO");
        exit(0);
      }
      req[cur] += k[nxt] * req[nxt];
    } else
      req[cur] += req[nxt];
  }
}
int main(int argc, char** argv) {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &b[i]);
  long long a;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a);
    req[i] = b[i] - a;
  }
  int x;
  for (int i = 2; i <= n; i++) {
    scanf("%d %lld", &x, &k[i]);
    G[x].push_back(i);
  }
  dfs(1);
  puts((req[1] < 0) ? "NO" : "YES");
  return 0;
}
