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<int> v[100009];
int parent[100009];
int s[100009];
int a[100009];
bool visited[100009];
void dfs(int node, int height, int sum) {
  visited[node] = true;
  if (height % 2 == 0) {
    int val = INT_MAX;
    for (int i = 0; i < v[node].size(); i++) {
      int nbr = v[node][i];
      if (!visited[nbr]) {
        val = min(val, s[nbr]);
      }
    }
    if (v[node].size() == 1) {
      a[node] = 0;
      s[node] = s[parent[node]];
    } else {
      int x = parent[node];
      val -= s[x];
      a[node] = val;
      s[node] = s[parent[node]] + a[node];
    }
  } else {
    a[node] = s[node] - sum;
  }
  for (int i = 0; i < v[node].size(); i++) {
    int nbr = v[node][i];
    if (!visited[nbr]) {
      dfs(nbr, height + 1, s[node]);
    }
  }
  return;
}
int main() {
  int n;
  cin >> n;
  for (int i = 2; i <= n; i++) cin >> parent[i];
  for (int i = 2; i <= n; i++) {
    int x = parent[i];
    v[i].push_back(x);
    v[x].push_back(i);
  }
  for (int i = 1; i <= n; i++) cin >> s[i];
  dfs(1, 1, 0);
  int flag = 0;
  long long int f = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] < 0) {
      flag = 1;
      break;
    }
    f += a[i];
  }
  if (flag == 1)
    cout << -1 << endl;
  else
    cout << f << endl;
  return 0;
}
