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;
int n, mus[100001], used[100001];
long long ans = 0;
bool mb = true;
vector<int> gr[100001];
void dfs(int edge, int parent, int deph) {
  int elem = 1000000001;
  if (deph % 2 == 0) {
    for (auto i : gr[edge]) {
      used[i] -= mus[parent];
      if (mus[i] < abs(used[i])) mb = false;
    }
    if (gr[edge].size() != 0) {
      for (auto i : gr[edge]) elem = min(mus[i] + used[i], elem);
      for (auto i : gr[edge]) used[i] -= elem;
      used[edge] = elem;
    }
  } else
    used[edge] += mus[edge];
  for (auto i : gr[edge]) dfs(i, edge, deph + 1);
}
int main() {
  cin >> n;
  int tmp;
  for (int i = 2; i <= n; i++) {
    cin >> tmp;
    gr[tmp].push_back(i);
  }
  for (int i = 0; i < n; i++) cin >> mus[i + 1];
  dfs(1, 1, 1);
  if (mb) {
    for (int i = 0; i < 100002; i++) {
      ans += used[i];
    }
    cout << ans;
  } else
    cout << -1;
  return 0;
}
