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 int P = 1000000007;
long long int a[100001], parent[100001], value[100001], flag = 0, ans = 0;
vector<long long int> adj[100001];
long long int visited[100001];
void dfs(long long int src, long long int val) {
  visited[src] = 1;
  for (long long int i = 0; i < adj[src].size(); i++) {
    if (!visited[adj[src][i]]) {
      if (value[adj[src][i]] < val && value[adj[src][i]] != -1) {
        flag = 1;
        return;
      } else {
        if (value[adj[src][i]] != -1) ans += value[adj[src][i]] - val;
        dfs(adj[src][i], max(value[adj[src][i]], val));
      }
    }
  }
}
int main() {
  long long int i, j, k, l, m, n, o, p, q, r, s, t, d;
  cin >> n;
  memset(visited, 0, sizeof(visited));
  for (i = 2; i <= n; i++) {
    cin >> parent[i];
    adj[parent[i]].push_back(i);
    adj[i].push_back(parent[i]);
  }
  for (i = 1; i <= n; i++) cin >> value[i];
  ans += value[1];
  long long int ma = value[1];
  for (i = 2; i <= n; i++) {
    long long int v1 = LLONG_MAX;
    if (value[i] == -1) {
      for (j = 0; j < adj[i].size(); j++) {
        if (parent[i] != adj[i][j]) {
          v1 = min(v1, value[adj[i][j]]);
        }
      }
      if (v1 != LLONG_MAX) value[i] = v1;
    }
  }
  dfs(1, ma);
  if (flag)
    cout << "-1\n";
  else
    cout << ans << "\n";
  return 0;
}
