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 a[200005], cost[200005], sum[200005], flag, ans;
vector<int> adj[200005];
vector<bool> visited(200005, false);
void BFS(int s) {
  queue<int> Q;
  Q.push(s);
  visited[s] = 1;
  cost[s] = a[s];
  sum[s] += a[s];
  ans += cost[s];
  while (!Q.empty()) {
    int u = Q.front();
    Q.pop();
    for (int i = 0; i < adj[u].size(); i++)
      if (visited[adj[u][i]] == 0) {
        int v = adj[u][i];
        if (a[v] == -1) {
          long long mn = 1e9;
          for (int j = 0; j < adj[v].size(); j++)
            if (visited[adj[v][j]] == 0) mn = min(mn, a[adj[v][j]] - sum[u]);
          if (mn == 1e9) mn = 0;
          a[v] = mn;
          for (int j = 0; j < adj[v].size(); j++)
            if (visited[adj[v][j]] == 0) {
              if (a[adj[v][j]] < sum[u]) flag = 1;
              a[adj[v][j]] -= (mn + sum[u]);
            }
        }
        cost[v] = a[v];
        sum[v] = cost[v] + sum[u];
        ans += cost[v];
        visited[v] = 1;
        Q.push(v);
      }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, u, v, s;
  cin >> n;
  for (int i = 2; i <= n; i++) {
    cin >> u;
    adj[u].push_back(i);
    adj[i].push_back(u);
  }
  for (int i = 1; i <= n; i++) {
    cin >> u;
    a[i] = u;
  }
  BFS(1);
  if (flag == 1)
    cout << -1 << endl;
  else
    cout << ans << endl;
  return 0;
}
