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<long long int> adj[100000 + 5];
long long int vis[100000 + 5] = {0}, a[100000 + 5] = {0};
long long int sp[100000 + 5], par[100000 + 5];
long long int n, den = 0, x;
void dfs(long long int s, long long int sum, long long int level) {
  long long int i, j, k;
  vis[s] = 1;
  for (i = 0; i < adj[s].size(); i++) {
    if (vis[adj[s][i]] == 0) {
      long long int val;
      par[adj[s][i]] = s;
      if (sp[adj[s][i]] == -1)
        val = 0;
      else {
        val = sp[adj[s][i]];
        if (val < sum) {
          den = 1;
          break;
        } else {
          if (level % 2 == 0) a[adj[s][i]] = val - sum;
        }
      }
      dfs(adj[s][i], max(sum, val), level + 1);
    }
  }
  if (level % 2 == 0) {
    long long int mn = 1000000000000;
    for (i = 0; i < adj[s].size(); i++) {
      if (par[s] != adj[s][i]) {
        mn = min(mn, a[adj[s][i]]);
      }
    }
    if (adj[s].size() >= 2) a[s] = mn;
    for (i = 0; i < adj[s].size(); i++) {
      if (par[s] != adj[s][i]) {
        a[adj[s][i]] = a[adj[s][i]] - mn;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int i, j, k;
  cin >> n;
  for (i = 2; i <= n; i++) {
    cin >> x;
    adj[x].push_back(i);
    adj[i].push_back(x);
  }
  for (i = 1; i <= n; i++) cin >> sp[i];
  a[1] = sp[1];
  dfs(1, sp[1], 1);
  if (den == 1)
    cout << "-1\n";
  else {
    long long int ans = 0;
    for (i = 1; i <= n; i++) ans += a[i];
    cout << ans << "\n";
  }
}
