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;
long long int max(long long int a, long long int b) {
  if (a > b)
    return a;
  else
    return b;
}
const int MAX = 100007;
vector<int> adj[MAX];
vector<int> s(MAX);
vector<int> par(MAX);
vector<int> ans(MAX);
vector<bool> visited(MAX, 0);
void dfs(int x, int dist) {
  if (visited[x]) return;
  visited[x] = true;
  if (dist % 2 == 0) {
    if (adj[x].empty()) {
      ans[x] = 0;
      return;
    }
    int p = par[x];
    int sp = s[p];
    int mn = 1e9 + 7;
    for (auto u : adj[x]) {
      mn = min(s[u] - sp, mn);
    }
    ans[x] = mn;
    s[x] = sp + ans[x];
  } else {
    ans[x] = s[x] - s[par[x]];
    if (x == 1) ans[x] = s[x];
  }
  for (auto u : adj[x]) {
    dfs(u, dist + 1);
  }
  return;
}
void solve() {
  int n;
  cin >> n;
  for (int i = 2; i <= n; i++) {
    cin >> par[i];
    adj[par[i]].push_back(i);
  }
  par[1] = 1;
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
  }
  dfs(1, 1);
  long long int tot = 0;
  bool flag = true;
  for (int i = 1; i <= n; i++) {
    tot += ans[i];
    if (ans[i] < 0) flag = false;
  }
  if (!flag) {
    cout << -1;
    return;
  }
  cout << tot;
}
int main() { solve(); }
