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<int> pr;
long long ans = 0;
void dfs(int v, vector<vector<int>> &g, vector<int> &s) {
  if (g[v].empty()) {
    if (s[v] == -1) {
      s[v] = s[pr[v]];
    }
  } else {
    if (s[v] == -1) {
      s[v] = s[g[v][0]];
      for (auto i : g[v]) {
        s[v] = min(s[v], s[i]);
        if (pr[v] != -1 && s[i] < s[pr[v]]) {
          cout << -1;
          exit(0);
        }
      }
      for (auto i : g[v]) dfs(i, g, s);
    } else {
      for (auto i : g[v]) {
        dfs(i, g, s);
      }
    }
  }
}
void dfs1(int v, vector<vector<int>> &g, vector<int> &s) {
  for (auto i : g[v]) {
    if (pr[i] != -1) ans += s[i] - s[pr[i]];
    dfs1(i, g, s);
  }
}
void solve() {
  int n;
  cin >> n;
  vector<int> s(n);
  pr.resize(n, -1);
  vector<vector<int>> g(n);
  for (int i = 1; i < n; i++) {
    int v;
    cin >> v;
    v--;
    g[v].push_back(i);
    pr[i] = v;
  }
  for (int i = 0; i < n; i++) cin >> s[i];
  vector<bool> used(n);
  dfs(0, g, s);
  dfs1(0, g, s);
  cout << ans + s[0];
}
int main() {
  solve();
  return 0;
}
