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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n;
  cin >> n;
  vector<vector<long long int>> v(n + 100);
  vector<long long int> v4(n + 100);
  v4[0] = -1;
  for (long long int i = 1; i < n; i++) {
    cin >> v4[i];
    v4[i] -= 1;
    v[v4[i]].push_back(i);
  }
  queue<pair<long long int, long long int>> q;
  q.push({0, 1});
  vector<long long int> v1(n, 0);
  for (long long int i = 0; i < n; i++) {
    cin >> v1[i];
  }
  vector<bool> visited(n + 100, false);
  visited[0] = true;
  bool val = true;
  long long int maxi = v1[0];
  vector<long long int> v2(n, 0);
  v2[0] = v1[0];
  while (!q.empty()) {
    pair<long long int, long long int> p = q.front();
    q.pop();
    long long int p1 = p.first;
    long long int p2 = p.second;
    long long int node = LONG_MAX;
    vector<long long int> v3;
    for (long long int i = 0; i < v[p1].size(); i++) {
      if (!visited[v[p1][i]]) {
        visited[v[p1][i]] = true;
        q.push({v[p1][i], p2 + (long long int)1});
        if ((p2 + 1) & 1) {
          if (v1[v4[p1]] > v1[v[p1][i]]) {
            val = false;
            break;
          }
          v3.push_back(v[p1][i]);
          node = min(node, v1[v[p1][i]] - v1[v4[p1]]);
        }
      }
    }
    if (p2 % 2 == 0 && v3.size()) {
      v2[p1] = node;
      for (long long int i = 0; i < v3.size(); i++) {
        v2[v3[i]] = v1[v3[i]] - v1[v4[p1]] - node;
      }
    }
  }
  if (!val) {
    cout << -1;
  } else {
    long long int sum = 0;
    for (long long int i = 0; i < v2.size(); i++) {
      sum += v2[i];
    }
    cout << sum;
  }
  return 0;
}
