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;
template <typename T>
inline bool chkmin(T &a, const T &b) {
  return b < a ? a = b, 1 : 0;
}
template <typename T>
inline bool chkmax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
const int oo = 0x3f3f3f3f;
const int maxn = 100100;
int n;
int fa[maxn + 5];
int s[maxn + 5];
int main() {
  scanf("%d", &n);
  fa[0] = -1;
  for (int i = (1), i_end_ = (n); i < i_end_; ++i) scanf("%d", fa + i), --fa[i];
  for (int i = (0), i_end_ = (n); i < i_end_; ++i) scanf("%d", s + i);
  for (int i = (0), i_end_ = (n); i < i_end_; ++i)
    if (s[i] == -1) s[i] = oo;
  long long ans = 0;
  for (int i = (0), i_end_ = (n); i < i_end_; ++i) {
    if (s[i] < oo) {
      chkmin(s[fa[i]], s[i]);
    }
  }
  for (int i = (0), i_end_ = (n); i < i_end_; ++i) {
    if (s[i] == oo) s[i] = s[fa[i]];
    if (s[fa[i]] > s[i]) {
      printf("-1\n");
      return 0;
    }
    ans += s[i];
    if (~fa[i]) ans -= s[fa[i]];
  }
  printf("%lld\n", ans);
  return 0;
}
