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 MAXN = 100000;
vector<vector<long long> > g(MAXN);
long long ans = 0;
long long n, p[100000], s[100000], m[100000];
int main() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    cin >> p[i];
    p[i]--;
    g[p[i]].push_back(i);
  }
  for (int i = 0; i < n; i++) {
    cin >> s[i];
  }
  for (int i = 0; i < n; i++) {
    if (s[i] == -1) {
      long long mn = -10000000, sm = 10000000;
      if (g[i].size() == 0)
        s[i] = s[p[i]];
      else {
        for (auto it : g[i]) sm = min(sm, s[it]);
        for (auto it : g[i]) {
          if (s[it] - s[p[i]] < 0) {
            cout << -1;
            return 0;
          }
          if (mn < s[it] - s[p[i]] and s[it] - s[p[i]] <= sm) {
            mn = s[it] - s[p[i]];
          }
        }
        s[i] = max(mn, sm);
      }
    }
    if (g[i].size() == 0)
      ans += s[i];
    else if (g[i].size() > 1) {
      for (int j = 0; j < g[i].size() - 1; j++) {
        ans -= s[i];
      }
    }
  }
  cout << ans;
}
