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>
#pragma GCC optimize("O3")
using namespace std;
int n;
vector<int> v[100002];
long long sum[100002];
long long val[100002];
void dfs(int dad, int nod) {
  if (sum[nod] != -1)
    val[nod] = sum[nod] - sum[dad];
  else {
    long long minsum = (1LL << 60);
    for (int j = 0; j < v[nod].size(); ++j)
      minsum = min(minsum, sum[v[nod][j]]);
    if (v[nod].size()) val[nod] = minsum - sum[dad];
    sum[nod] = sum[dad] + val[nod];
  }
  for (int j = 0; j < v[nod].size(); ++j) dfs(nod, v[nod][j]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 2; i <= n; ++i) {
    int x;
    cin >> x;
    v[x].push_back(i);
  }
  for (int i = 1; i <= n; ++i) cin >> sum[i];
  dfs(0, 1);
  long long ans = 0;
  for (int j = 1; j <= n; ++j)
    if (val[j] < 0) {
      cout << -1;
      return 0;
    } else
      ans += val[j];
  cout << ans;
  return 0;
}
