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<vector<int> > g;
int flag = 1;
long long a[100001];
long long ans[100001];
void dfs(int x, long long s) {
  if (a[x] != -1) {
    if (a[x] - s < 0) flag = 0;
    ans[x] = a[x] - s;
    for (int i = 0; i < g[x].size(); i++) {
      int y = g[x][i];
      dfs(y, a[x]);
    }
  } else {
    if (g[x].size() == 0) return;
    ans[x] = 1000000000;
    for (int i = 0; i < g[x].size(); i++) {
      int y = g[x][i];
      ans[x] = min(ans[x], a[y] - s);
      if (a[y] - s < 0) flag = 0;
    }
    for (int i = 0; i < g[x].size(); i++) {
      int y = g[x][i];
      dfs(y, ans[x] + s);
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  g.resize(n + 1);
  for (int i = 0; i < n - 1; i++) {
    int b;
    cin >> b;
    g[b].push_back(i + 2);
  }
  for (int i = 1; i <= n; i++) cin >> a[i];
  dfs(1, 0);
  long long an = 0;
  if (flag == 0) {
    cout << -1 << endl;
    return 0;
  }
  for (int i = 1; i <= n; i++) an += ans[i];
  cout << an << endl;
}
