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 s[100020];
long long ans;
vector<int> G[100020];
void dfs(int n, int now) {
  if (s[n] == -1) {
    for (int i = 0; i < G[n].size(); i++) {
      dfs(G[n][i], 0);
      if (s[n] == -1)
        s[n] = s[G[n][i]];
      else
        s[n] = min(s[n], s[G[n][i]]);
    }
    if (s[n] == -1) s[n] = now;
  } else {
    for (int i = 0; i < G[n].size(); i++) {
      dfs(G[n][i], s[n]);
      if (s[n] > s[G[n][i]]) {
        cout << "-1";
        exit(0);
      }
    }
  }
  for (int i = 0; i < G[n].size(); i++) ans += s[G[n][i]] - s[n];
}
int main() {
  ios::sync_with_stdio(false);
  int n, k;
  cin >> n;
  for (int i = 2; i <= n; i++) {
    cin >> k;
    G[k].push_back(i);
  }
  for (int i = 1; i <= n; i++) cin >> s[i];
  ans = s[1];
  dfs(1, 0);
  cout << ans;
  return 0;
}
