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;
const long long INF = 1e9 + 1;
const int N = 1e5 + 5;
int n, p[N], a[N], t[N], d[N], s[N];
vector<int> adj[N];
long long res;
void Input() {
  cin >> n;
  for (int i = 2; i <= n; i++) {
    int p;
    cin >> p;
    adj[p].push_back(i);
    adj[i].push_back(p);
  }
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
    t[i] = INF;
  }
  a[1] = s[1];
}
void DFS(int u, int pre) {
  for (int v : adj[u]) {
    if (v == pre) continue;
    p[v] = u, d[v] = 1 - d[u];
    DFS(v, u);
    if (s[u] < 0 && adj[u].size() > 2) t[u] = min(t[u], s[v]);
  }
}
void Get(int u, int pre, int sum) {
  if (s[u] < 0 && t[u] < INF) a[u] = t[u] - sum, sum = t[u];
  if (s[u] != -1 && u > 1) a[u] = s[u] - sum, sum = s[u];
  for (int v : adj[u]) {
    if (v == pre) continue;
    Get(v, u, sum);
  }
}
void Solve() {
  d[1] = 1, DFS(1, 0);
  for (int i = 2; i <= n; i++) {
    if (d[i] && d[p[p[i]]] && s[i] < s[p[p[i]]]) {
      cout << -1 << '\n';
      return;
    }
  }
  Get(1, 0, a[1]);
  for (int i = 1; i <= n; i++) res += a[i];
  cout << res << '\n';
}
int main() {
  Input();
  Solve();
  return 0;
}
