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 int MAXN = 100010;
const long long MAXELE = numeric_limits<long long>::max();
vector<int> arv(MAXN), h(MAXN);
vector<long long> s(MAXN);
vector<long long> val(MAXN, MAXELE);
vector<vector<int> > vecH(MAXN + 10);
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  arv[0] = 0;
  h[0] = 1;
  vecH[1].push_back(0);
  for (int i = (1); i < (int)(n); ++i) {
    int a;
    cin >> a;
    a--;
    arv[i] = a;
  }
  for (int i = 0; i < (int)(n); ++i) cin >> s[i];
  val[0] = s[0];
  for (int i = (1); i < (int)(n); ++i) {
    int at = i;
    while (h[at] == 0) {
      at = arv[at];
    }
    h[i] = h[at] + 1;
    vecH[h[i]].push_back(i);
    ;
    ;
  }
  int maxH = *max_element((h).begin(), (h).end());
  ;
  for (int i = 3; i <= maxH; i += 2) {
    for (int v : vecH[i]) {
      if (s[v] >= s[arv[arv[v]]]) {
        long long diff = s[v] - s[arv[arv[v]]];
        val[arv[v]] = min(val[arv[v]], diff);
      } else {
        cout << -1 << '\n';
        return 0;
      }
    }
    for (int v : vecH[i]) {
      val[v] = s[v] - s[arv[arv[v]]] - val[arv[v]];
    }
  }
  for (int i = 2; i <= maxH; i += 2) {
    for (int v : vecH[i]) {
      if (val[v] == MAXELE) val[v] = 0;
    }
  }
  long long sum = 0;
  for (int i = 0; i < (int)(n); ++i) sum += val[i];
  cout << sum << '\n';
  return 0;
}
