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;
void bfs(int p[], int s[], int n, vector<vector<int> > adj) {
  queue<int> q;
  q.push(1);
  int t = 0;
  int mino = INT_MAX;
  bool vis[n + 1];
  for (int i = 1; i < n + 1; ++i) {
    vis[i] = false;
  }
  vis[1] = true;
  while (!q.empty()) {
    int t = q.front();
    q.pop();
    vis[t] = true;
    if (s[t] == -1) {
      for (int x : adj[t]) {
        mino = min(s[x], mino);
        if (!vis[x]) q.push(x);
      }
      s[t] = mino;
      mino = INT_MAX;
      if (adj[t].size() == 0) s[t] = s[p[t]];
    } else
      for (int x : adj[t]) q.push(x);
  }
  int a[n + 1];
  a[1] = s[1];
  long long int ans = a[1];
  for (int i = 2; i < n + 1; i++) {
    a[i] = s[i] - s[p[i]];
    if (a[i] < 0) {
      cout << "-1" << endl;
      return;
    }
    ans += a[i];
  }
  cout << ans << endl;
}
int main() {
  int n;
  cin >> n;
  int p[n + 1];
  vector<vector<int> > adj(n + 1);
  int x;
  p[1] = p[0] = -1;
  for (int i = 2; i <= n; i++) {
    cin >> p[i];
    adj[p[i]].push_back(i);
  }
  int s[n + 1];
  s[0] = -1;
  for (int i = 1; i < n + 1; ++i) cin >> s[i];
  bfs(p, s, n, adj);
  return 0;
}
