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<int> adj[100005];
int val[100005], par[100005];
int ans[100005];
void dfs(int x = 1, int dep = 1) {
  if (x == 1)
    ans[x] = val[x];
  else if (val[x] == -1)
    ans[x] = 0, val[x] = val[par[x]];
  else
    ans[x] = val[x] - val[par[x]];
  int mn = 1000000005;
  for (int i = 0; i < adj[x].size(); i++) {
    dfs(adj[x][i], dep + 1);
    mn = min(mn, ans[adj[x][i]]);
  }
  if (dep % 2 == 0 && mn != 1000000005) {
    ans[x] += mn;
    val[x] += mn;
    for (int i = 0; i < adj[x].size(); i++) {
      ans[adj[x][i]] -= mn;
    }
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 2; i <= n; i++) {
    int x;
    cin >> x;
    adj[x].push_back(i);
    par[i] = x;
  }
  for (int i = 1; i <= n; i++) cin >> val[i];
  dfs();
  long long int ansr = 0;
  for (int i = 1; i <= n; i++) {
    if (ans[i] < 0) {
      cout << "-1" << endl;
      return 0;
    }
    ansr += ans[i];
  }
  cout << ansr << endl;
  return 0;
}
