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>
#pragma GCC optimize("O3")
using namespace std;
const long long mod = 1ll * 4 * 1024 * 1024 * 1024;
const long long inf = 1e9 + 7;
bool b = false;
void dfs(int u, int p, long long sum, vector<bool> &used, vector<long long> &a,
         vector<long long> &ans, vector<vector<long long> > &g) {
  used[u] = true;
  if (p == -1) {
    sum = a[0];
    ans[0] = a[0];
  } else {
    if (a[u] != -1) {
      ans[u] = a[u] - sum;
      sum = a[u];
    } else {
      if (g[u].empty()) {
        ans[u] = 0;
        return;
      }
      long long mini = 1e9;
      for (auto v : g[u]) mini = min(mini, a[v]);
      ans[u] = mini - sum;
      sum += ans[u];
    }
  }
  for (auto v : g[u])
    if (!used[v]) dfs(v, u, sum, used, a, ans, g);
}
int main() {
  int n;
  cin >> n;
  vector<vector<long long> > g(n);
  for (int i = 1; i < n; ++i) {
    int j;
    cin >> j;
    j--;
    g[j].push_back(i);
  }
  vector<long long> a(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  vector<bool> used(n, false);
  vector<long long> ans(n, 1e9);
  ans[0] = a[0];
  dfs(0, -1, 0, used, a, ans, g);
  long long sum = 0;
  for (auto x : ans) {
    sum += x;
    b |= (x < 0);
  }
  if (b) {
    cout << -1;
    return 0;
  }
  cout << sum;
  return 0;
}
