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;
const int maxN = 1e5 + 100;
const int inf = 1e9 + 100;
int n;
class NODE {
 public:
  int par;
  vector<int> adj;
  int sumAll;
  int sum2;
  int depth;
  NODE() {}
} nodes[maxN];
long long ans = 0;
void dfs(int u, int depth) {
  int MIN = inf;
  nodes[u].depth = depth;
  if (depth % 2 == 0 && u != 0) {
    nodes[u].sum2 = nodes[u].sumAll - nodes[nodes[nodes[u].par].par].sumAll;
    if (nodes[u].sum2 < 0) {
      cout << -1;
      exit(0);
    }
  }
  for (int i = 0; i < nodes[u].adj.size(); i++) {
    int v = nodes[u].adj[i];
    dfs(v, depth + 1);
    MIN = min(MIN, nodes[v].sum2);
  }
  if (depth % 2 == 1) {
    if (MIN != inf) ans += MIN;
    for (int i = 0; i < nodes[u].adj.size(); i++) {
      int v = nodes[u].adj[i];
      ans += (nodes[v].sum2 - MIN);
    }
  }
}
int main() {
  iostream::sync_with_stdio(0);
  cin >> n;
  nodes[0].par = -1;
  for (int i = 1; i < n; i++) {
    int par;
    cin >> par;
    par--;
    nodes[i].par = par;
    nodes[par].adj.push_back(i);
  }
  for (int i = 0; i < n; i++) {
    int sum;
    cin >> sum;
    nodes[i].sumAll = sum;
  }
  dfs(0, 0);
  cout << ans + nodes[0].sumAll;
}
