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;
long long n;
vector<vector<long long> > adj;
vector<long long> a, s;
bool flag = false;
void DFS(long long curr, long long par, long long level) {
  if (flag) return;
  if (level % 2 == 0) {
    long long choice = INT64_MAX;
    for (auto x : adj[curr]) {
      if (x == par) continue;
      choice = min(choice, s[x]);
    }
    if (choice == INT64_MAX) {
      a[curr] = 0;
      s[curr] = a[curr] + s[par];
    } else {
      s[curr] = choice;
      a[curr] = s[curr] - s[par];
      if (a[curr] < 0) {
        flag = true;
        return;
      }
    }
  } else {
    if (par == -1) {
      a[curr] = s[curr];
    } else {
      a[curr] = s[curr] - s[par];
      if (a[curr < 0]) {
        flag = true;
        return;
      }
    }
  }
  for (auto x : adj[curr]) {
    if (x == par) continue;
    DFS(x, curr, level + 1);
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  adj.resize(n + 1);
  a.resize(n + 1);
  s.resize(n + 1);
  for (long long i = 2; i <= n; i++) {
    long long p;
    cin >> p;
    adj[i].push_back(p);
    adj[p].push_back(i);
  }
  for (long long i = 1; i <= n; i++) {
    cin >> s[i];
  }
  DFS(1, -1, 1);
  if (flag) {
    cout << "-1";
  } else {
    long long sum = 0;
    for (long long i = 1; i <= n; i++) {
      sum = sum + a[i];
    }
    cout << sum;
  }
  return 0;
}
