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 MAX_N = 1e5 + 5;
int par[MAX_N];
int lvl[MAX_N];
int S[MAX_N];
int A[MAX_N];
vector<int> adj[MAX_N];
int main() {
  ios::sync_with_stdio(false);
  int vertexc;
  cin >> vertexc;
  lvl[1] = 1;
  for (int i = 2; i <= vertexc; i++) {
    cin >> par[i];
    lvl[i] = lvl[par[i]] + 1;
    adj[par[i]].push_back(i);
  }
  for (int i = 1; i <= vertexc; i++) {
    cin >> S[i];
  }
  A[1] = S[1];
  for (int i = 2; i <= vertexc; i++) {
    if (lvl[i] % 2 == 0 && S[i] != -1) {
      cout << -1 << endl;
      return 0;
    }
    if (lvl[i] % 2 != 0 && S[i] == -1) {
      cout << -1 << endl;
      return 0;
    }
    if (lvl[i] % 2 != 0) {
      A[i] = S[i] - S[par[par[i]]];
      if (A[i] < 0) {
        cout << -1 << endl;
        return 0;
      }
    }
  }
  for (int i = 2; i <= vertexc; i++) {
    if (lvl[i] % 2 == 0) {
      A[i] = 1 << 30;
      if (adj[i].empty()) {
        A[i] = 0;
      }
      for (int nxt : adj[i]) {
        A[i] = min(A[i], A[nxt]);
      }
      for (int nxt : adj[i]) {
        A[nxt] -= A[i];
      }
    }
  }
  long long ans = 0;
  for (int i = 1; i <= vertexc; i++) {
    ans += A[i];
  }
  cout << ans << endl;
}
