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;
struct Node {
  int parent;
  vector<int> adj;
  int val;
  int sum;
};
struct Node nodes[100005];
bool vis1[100005];
bool vis2[100005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  memset(vis1, false, sizeof vis1);
  memset(vis2, false, sizeof vis2);
  cin >> n;
  for (int i = 2; i <= n; i++) {
    int a;
    cin >> a;
    nodes[i].parent = a;
    nodes[a].adj.push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    int a;
    cin >> a;
    nodes[i].sum = a;
    nodes[i].val = a;
    if (i == 1) {
      nodes[i].parent = -1;
    }
  }
  queue<int> q;
  bool valid = true;
  vis1[1] = 1;
  q.push(1);
  while (!q.empty()) {
    int d = q.front();
    q.pop();
    Node front = nodes[d];
    int parent = front.parent;
    if (front.val == -1) {
      int min = 1500000000;
      int parentval = nodes[parent].sum;
      for (auto x : front.adj) {
        if (!vis1[x]) {
          vis1[x] = 1;
          if (nodes[x].sum - parentval >= 0) {
            if (nodes[x].sum - parentval <= min) {
              min = nodes[x].sum - parentval;
            }
          } else {
            valid = false;
            break;
          }
          q.push(x);
        }
      }
      nodes[d].sum = min + nodes[parent].sum;
      nodes[d].val = min;
      if (front.adj.size() == 0) {
        nodes[d].val = 0;
      }
    } else {
      if (front.parent != -1) {
        Node parentObject = nodes[parent];
        Node grandParent = nodes[parentObject.parent];
        nodes[d].val = nodes[d].sum - parentObject.val - grandParent.sum;
      }
      for (auto x : front.adj) {
        if (!vis1[x]) {
          vis1[x] = 1;
          q.push(x);
        }
      }
    }
  }
  long long sum = 0;
  if (valid) {
    for (int i = 1; i <= n; i++) {
      sum += (long long)nodes[i].val;
    }
    cout << sum << '\n';
  } else {
    cout << -1 << '\n';
  }
  return 0;
}
