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;
template <class T>
void printQueue(T& q) {
  while (!q.empty()) {
    cout << q.top() << endl;
    q.pop();
  }
}
vector<int> adj[100009];
long long values[100009];
long long sum;
long long ans[100009];
int paradox;
bool visited[100009];
void dfs(int s, long long exp) {
  visited[s] = true;
  if (values[s] == -1) {
    long long astha = 1e18;
    for (int i = 0; i < adj[s].size(); i++) {
      if (visited[adj[s][i]] == false) {
        astha = min(astha, values[adj[s][i]]);
      }
    }
    if (astha == 1e18) {
      ans[s] = 0;
    } else {
      ans[s] = astha - exp;
    }
    if (ans[s] < 0) {
      paradox = 1;
    }
    sum = sum + ans[s];
  } else {
    ans[s] = values[s] - exp;
    if (ans[s] < 0) {
      paradox = 1;
    }
    sum = sum + ans[s];
  }
  for (int i = 0; i < adj[s].size(); i++) {
    if (visited[adj[s][i]] == false) {
      dfs(adj[s][i], exp + ans[s]);
    }
  }
}
int main() {
  int t = 1;
  paradox = 0;
  sum = 0;
  memset(visited, false, sizeof(visited));
  while (t--) {
    int n, x;
    cin >> n;
    for (int i = 2; i <= n; i++) {
      cin >> x;
      adj[x].push_back(i);
      adj[i].push_back(x);
    }
    for (int i = 1; i <= n; i++) {
      cin >> values[i];
    }
    dfs(1, 0);
    if (paradox == 1) {
      cout << -1 << endl;
    } else {
      cout << sum << endl;
    }
  }
}
