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, p, h[100005], par[100005];
long long ans, s[100005];
vector<int> a[100005];
int main() {
  cin >> n;
  h[1] = 1;
  for (int i = 2; i <= n; i++) {
    cin >> p;
    par[i] = p;
    h[i] = h[p] + 1;
    a[p].push_back(i);
  }
  for (int i = 1; i <= n; i++) cin >> s[i];
  ans = 0;
  for (int u = 1; u <= n; u++) {
    if (h[u] % 2 == 0) {
      s[u] = 1000000000;
      for (int v : a[u]) s[u] = min(s[u], s[v]);
      if (s[u] == 1000000000) s[u] = s[par[u]];
      if (s[u] < s[par[u]]) return cout << -1, 0;
    }
    ans += s[u] - s[par[u]];
  }
  cout << ans;
  return 0;
}
