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;
int par[100005];
int n;
int depth[100005];
int s[100005];
int a[100005];
vector<vector<int> > graph(100005, vector<int>());
void dfs(int v, int prv) {
  if (s[v] == -1) {
    int minval = 1000000007;
    for (int u : graph[v]) {
      if (s[u] == -1) {
        cout << -1;
        exit(0);
      }
      minval = min(minval, s[u]);
    }
    if (minval == 1000000007) {
      minval = prv;
    }
    a[v] = minval - prv;
    if (a[v] < 0) {
      cout << -1;
      exit(0);
    }
    for (int u : graph[v]) {
      dfs(u, a[v] + prv);
    }
  } else {
    a[v] = s[v] - prv;
    if (a[v] < 0) {
      cout << -1;
      exit(0);
    }
    for (int u : graph[v]) {
      if (s[u] != -1) {
        cout << -1;
        exit(0);
      } else {
        dfs(u, s[v]);
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 2; i <= n; i++) {
    cin >> par[i];
    graph[par[i]].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
  }
  dfs(1, 0);
  long long int ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += a[i];
  }
  cout << ans;
  return 0;
}
