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;
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m, x = 0, z = 0;
  cin >> n;
  vector<long long> a(n), s(n), pr(n);
  vector<vector<long long>> dety(n);
  for (long long j = 1; j < n; j++) {
    cin >> x;
    pr[j] = x - 1;
    dety[pr[j]].push_back(j);
  }
  for (long long j = 0; j < n; j++) {
    cin >> s[j];
  }
  a[0] = s[0];
  queue<long long> q;
  q.push(0);
  while (!q.empty()) {
    long long v = q.front();
    q.pop();
    long long mi = 1000000000000;
    for (long long u : dety[v]) {
      q.push(u);
      if (s[u] < mi) {
        mi = s[u];
      };
    }
    if (s[v] == -1) {
      if (mi == 1000000000000) {
        mi = s[pr[v]];
      }
      s[v] = mi;
      a[v] = s[v] - s[pr[v]];
      if (s[pr[v]] > s[v]) {
        cout << -1;
        return 0;
      }
      for (long long u : dety[v]) {
        a[u] = s[u] - s[v];
      }
    }
  }
  long long otv = 0;
  for (long long j = 0; j < n; j++) {
    otv += a[j];
  }
  cout << otv;
}
