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> childs[100000 + 1];
long long s[100000 + 1];
int p[100000 + 1];
long long a[100000 + 1];
bool flag = false;
void fill(int v, long long prevsum) {
  if (s[v] != -1) {
    a[v] = s[v] - prevsum;
    if (a[v] < 0) flag = true;
    for (int i = 0; i < childs[v].size(); ++i) {
      fill(childs[v][i], s[v]);
    }
  } else {
    long long mi = 1000000001;
    for (int i = 0; i < childs[v].size(); ++i) {
      mi = min(mi, s[childs[v][i]]);
    }
    if (mi == 1000000001) mi = prevsum;
    a[v] = mi - prevsum;
    if (a[v] < 0) flag = true;
    for (int i = 0; i < childs[v].size(); ++i) {
      fill(childs[v][i], mi);
    }
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    childs[i].clear();
  }
  for (int i = 2; i < n + 1; ++i) {
    cin >> p[i];
    childs[p[i]].push_back(i);
  }
  for (int i = 1; i < n + 1; ++i) {
    cin >> s[i];
  }
  fill(1, 0);
  if (flag)
    cout << -1 << endl;
  else {
    long long sum = 0;
    for (int i = 1; i < n + 1; ++i) {
      sum += a[i];
    }
    cout << sum << endl;
  }
}
