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<long long> > adj(100005);
long long sum[100005], arr[100005], ans, f = 0;
void dfs(long long curr, long long par, long long flag) {
  if (flag == 1) {
    for (int i = 0; i < (long long)adj[curr].size(); i++) {
      dfs(adj[curr][i], curr, 1 - flag);
    }
  } else {
    long long avail = sum[par];
    long long mn = 1e15;
    for (int i = 0; i < (long long)adj[curr].size(); i++) {
      mn = min(mn, sum[adj[curr][i]]);
    }
    if (mn != 1e15) {
      if (mn < avail) {
        f = 1;
      }
      mn -= avail;
      ans += mn;
      avail += mn;
    }
    for (int i = 0; i < (long long)adj[curr].size(); i++) {
      ans += sum[adj[curr][i]] - avail;
      dfs(adj[curr][i], curr, 1 - flag);
    }
  }
}
int main() {
  long long n;
  cin >> n;
  for (int i = 2; i <= n; i++) {
    long long par;
    cin >> par;
    adj[par].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    cin >> sum[i];
  }
  ans = sum[1];
  dfs(1, -1, 1);
  if (f == 1) {
    cout << "-1";
    return 0;
  }
  cout << ans << "\n";
}
