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 mxn = 1e5 + 5;
long long S[mxn], a[mxn], N;
vector<vector<long long> > g;
long long upd(long long parent, long long node, long long al) {
  long long tmp = LONG_MAX;
  for (auto u : g[node]) {
    if (u != parent) tmp = min(tmp, upd(node, u, al + 1) - S[parent]);
  }
  if (al % 2 == 0) {
    a[node] = (tmp == LONG_MAX) ? 0 : tmp;
    S[node] = S[parent] + a[node];
  }
  return S[node];
}
void odd(long long parent, long long node, long long h) {
  if (S[node] < S[parent]) {
    cout << "-1";
    exit(0);
  }
  if (h % 2 == 1) {
    a[node] = S[node] - S[parent];
  }
  for (auto u : g[node]) {
    if (u != parent) {
      odd(node, u, h + 1);
    }
  }
}
int main() {
  scanf("%d", &N);
  g.resize(N + 3);
  for (long long i = 2; i <= N; i++) {
    long long a;
    cin >> a;
    g[a].push_back(i);
  }
  for (long long i = 1; i <= N; i++) {
    cin >> S[i];
  }
  upd(0, 1, 1);
  odd(0, 1, 1);
  long long ans = 0;
  for (int i = 1; i <= N; i++) {
    ans += a[i];
  }
  cout << ans;
  return 0;
}
