Mitya has a rooted tree with n vertices indexed from 1 to n, where the root has index 1. Each vertex v initially had an integer number a_v ≥ 0 written on it. For every vertex v Mitya has computed s_v: the sum of all values written on the vertices on the path from vertex v to the root, as well as h_v — the depth of vertex v, which denotes the number of vertices on the path from vertex v to the root. Clearly, s_1=a_1 and h_1=1.

Then Mitya erased all numbers a_v, and by accident he also erased all values s_v for vertices with even depth (vertices with even h_v). Your task is to restore the values a_v for every vertex, or determine that Mitya made a mistake. In case there are multiple ways to restore the values, you're required to find one which minimizes the total sum of values a_v for all vertices in the tree.

Input

The first line contains one integer n — the number of vertices in the tree (2 ≤ n ≤ 10^5). The following line contains integers p_2, p_3, ... p_n, where p_i stands for the parent of vertex with index i in the tree (1 ≤ p_i < i). The last line contains integer values s_1, s_2, ..., s_n (-1 ≤ s_v ≤ 10^9), where erased values are replaced by -1.

Output

Output one integer — the minimum total sum of all values a_v in the original tree, or -1 if such tree does not exist.

Examples

Input


5
1 1 1 1
1 -1 -1 -1 -1


Output


1


Input


5
1 2 3 1
1 -1 2 -1 -1


Output


2


Input


3
1 2
2 -1 1


Output


-1
#include <bits/stdc++.h>
const int mod = 1000000007;
const int inf = 1000000009;
const double eps = 1e-6;
const long long int INF = 1000000000000000009;
using namespace std;
vector<int> P[100005];
int T[100005];
int D[100005];
bool O[100005];
int F[100005];
void bfs() {
  queue<int> Q;
  Q.push(1);
  O[1] = true;
  while (!Q.empty()) {
    int k = Q.front();
    Q.pop();
    int mini = inf;
    for (auto v : P[k]) {
      if (!O[v]) {
        O[v] = true;
        F[v] = k;
        mini = min(mini, D[v] - D[F[k]]);
        Q.push(v);
      }
    }
    if (D[k] == -1) {
      if (mini == inf) mini = 0;
      T[k] = mini;
      D[k] = mini + D[F[k]];
    } else {
      T[k] = D[k] - D[F[k]];
    }
  }
}
long long int wynik = 0;
bool czy = true;
long long int sum = 0;
void check(int n, int p) {
  sum += T[n];
  wynik += T[n];
  if (T[n] < 0) czy = false;
  if (sum != D[n]) czy = false;
  for (auto v : P[n]) {
    if (v != p) {
      check(v, n);
    }
  }
  sum -= T[n];
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    int a;
    scanf("%d", &a);
    P[a].push_back(i);
    P[i].push_back(a);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &D[i]);
  }
  bfs();
  check(1, 0);
  if (!czy) {
    printf("-1");
    return 0;
  }
  printf("%lld", wynik);
  return 0;
}
