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 long long M = 998244353;
const long long oo = 1e18;
struct tree {
  long long sum = 0, mx = 0, ro = 0;
  vector<int> n;
} s[100010];
long long n, p[100010], s1[100010], s2[100010];
void check(int v, int sm) {
  if (s1[v] == -1) {
    s1[v] = sm;
  } else {
    if (sm > s1[v]) {
      cout << -1 << endl;
      exit(0);
    }
    s2[v] = s1[v] - sm;
  }
  for (auto x : s[v].n) {
    check(x, s1[v]);
  }
}
int main() {
  cin >> n;
  for (int i = 2; i <= n; ++i) {
    cin >> p[i];
    s[p[i]].n.push_back(i);
  }
  for (int i = 1; i <= n; ++i) {
    cin >> s1[i];
    s[i].sum = s1[i];
    s[i].ro = s1[i];
  }
  check(1, 0);
  for (int i = n; i > 0; --i) {
    if (s[i].ro == -1) {
      if (!s[i].n.size()) continue;
      long long mn = oo;
      for (auto x : s[i].n) {
        mn = min(mn, s2[x]);
      }
      s2[i] += mn;
      for (auto x : s[i].n) {
        s2[x] -= mn;
      }
    }
  }
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    ans += s2[i];
  }
  cout << ans << endl;
}
