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;
vector<int> ch[100005];
long long s[100005];
int p[100005];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 2; i <= n; ++i) {
    scanf("%d", &p[i]);
    ch[p[i]].push_back(i);
  }
  for (int i = 1; i <= n; ++i) scanf("%lld", &s[i]);
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    if (s[i] == -1) {
      long long v = 1LL << 60;
      for (int x : ch[i]) {
        v = min(v, s[x] - s[p[i]]);
      }
      if (v < 0) {
        printf("-1\n");
        return 0;
      }
      if (ch[i].empty()) v = 0;
      s[i] = v + s[p[i]];
      ans += v;
    } else
      ans += s[i] - s[p[i]];
  }
  printf("%lld\n", ans);
}
