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;
long long n, p[100005], sum[100005], t, ans, mn;
queue<long long> q;
vector<long long> v[100005];
int main() {
  scanf("%lld", &n);
  for (int i = 2; i <= n; i++) {
    scanf("%lld", &p[i]);
    v[p[i]].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &sum[i]);
  }
  q.push(1);
  while (q.size()) {
    t = q.front();
    q.pop();
    if (sum[t] == -1) {
      if (!v[t].size()) {
        goto skip;
      }
      mn = 2147483640;
      for (int i = 0; i < v[t].size(); i++) {
        mn = min(mn, sum[v[t][i]]);
      }
      if (mn < sum[p[t]]) {
        printf("-1\n");
        return 0;
      }
      sum[t] = mn;
      ans += (sum[t] - sum[p[t]]);
    skip:;
    } else {
      if (sum[t] < sum[p[t]]) {
        printf("-1\n");
        return 0;
      }
      ans += (sum[t] - sum[p[t]]);
    }
    for (int i = 0; i < v[t].size(); i++) {
      q.push(v[t][i]);
    }
  }
  printf("%lld\n", ans);
  return 0;
}
