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>
using namespace std;
vector<long long int> adj[100005];
long long int A[100005], flag;
void fun(long long int x, long long int v, long long int h) {
  if (flag == 1) return;
  if (adj[x].size() == 0) {
    if (h % 2 == 0) {
      A[x] = 0;
    } else {
      if (A[x] < v) flag = 1;
      A[x] = A[x] - v;
    }
    return;
  }
  long long int V, M = 1000000009;
  if (h % 2 == 0) {
    for (int i = 0; i < adj[x].size(); i++) {
      M = min(M, A[adj[x][i]]);
    }
    if (M < v) flag = 1;
    A[x] = M - v;
    V = M;
  } else {
    V = A[x];
    A[x] = A[x] - v;
    if (A[x] < 0) flag = 1;
  }
  for (int i = 0; i < adj[x].size(); i++) {
    fun(adj[x][i], V, h + 1);
  }
}
int main() {
  long long int n, a;
  cin >> n;
  for (int i = 2; i <= n; i++) {
    cin >> a;
    adj[a].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    cin >> A[i];
  }
  fun(1, 0, 1);
  if (flag == 1)
    cout << -1;
  else {
    long long int S = 0;
    for (int i = 1; i <= n; i++) S += A[i];
    cout << S;
  }
  return 0;
}
