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;
vector<long long int> g[200005];
long long int val[200005];
long long int ans = 0;
bool flag = true;
void dfs(long long int src, long long int par) {
  bool iseven = false;
  long long int k;
  if (par != -1) k = val[par];
  long long int mn = 1e10, cnt = 0, curr = 0;
  if (val[src] == -1) {
    iseven = true;
  }
  for (auto x : g[src]) {
    if (x == par) continue;
    dfs(x, src);
    if (iseven) {
      mn = min(mn, val[x]);
      curr += val[x];
      cnt++;
    }
  }
  if (iseven && cnt > 0) {
    if (mn < k) {
      flag = false;
      return;
    }
    ans += (curr - (cnt - 1) * (mn - k) - k * cnt);
  }
}
signed main() {
  long long int n;
  cin >> n;
  for (long long int i = 2; i <= n; i++) {
    long long int x;
    cin >> x;
    g[x].push_back(i);
    g[i].push_back(x);
  }
  for (long long int i = 1; i <= n; i++) cin >> val[i];
  dfs(1, -1);
  if (flag)
    cout << ans + val[1] << endl;
  else
    cout << -1 << endl;
}
