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 int N = 1e5 + 7;
int n;
long long ans = 0;
bool vis[N];
vector<int> adj[N];
int c[N];
bool f = 0;
void dfs(int i, int y) {
  vis[i] = 1;
  if (c[i] == -1) {
    if (adj[i].size() > 0) {
      long t = 1e9 + 43;
      for (auto x : adj[i]) {
        if (c[x] < t) {
          t = c[x];
        }
      }
      if (t < y) f = 1;
      ans += t - y;
      y = t;
    }
  } else {
    if (c[i] >= y) {
      ans += c[i] - y;
      y = c[i];
    } else {
      f = 1;
    }
  }
  for (auto v : adj[i]) {
    if (!vis[v]) {
      dfs(v, y);
    }
  }
}
int main() {
  cin >> n;
  int u;
  for (int i = 2; i <= n; i++) {
    cin >> u;
    adj[u].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    cin >> c[i];
  }
  dfs(1, 0);
  if (f) {
    cout << -1;
    return 0;
  }
  cout << ans;
}
