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 unsigned long long int mod = 1e9 + 7;
const int N = (1e5);
vector<set<pair<int, int> > > v(N + 1);
int a[N + 1] = {0};
bool ans = true;
int s[N + 1];
bool vist[N + 1];
void dfs(int src, int fsum, int par) {
  if (vist[src]) return;
  vist[src] = true;
  if (s[src] == -1) {
    auto it = *v[src].begin();
    if (it.second == par) v[src].erase(it);
    if (!v[src].empty()) {
      it = *v[src].begin();
      a[src] = s[it.second] - fsum;
    }
    s[src] = a[src] + fsum;
  } else {
    a[src] = s[src] - fsum;
  }
  if (a[src] < 0) {
    ans = false;
  }
  for (auto i : v[src]) {
    if (!vist[i.second]) {
      dfs(i.second, s[src], src);
    }
  }
}
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  int p[n + 1];
  for (int i = 2; i <= n; i++) cin >> p[i];
  for (int i = 1; i <= n; i++) cin >> s[i];
  for (int i = 2; i <= n; i++) {
    v[i].insert({s[p[i]], p[i]});
    v[p[i]].insert({s[i], i});
  }
  dfs(1, 0, 0);
  unsigned long long int sum = 0;
  for (int i = 1; i <= n; i++) {
    sum += a[i];
  }
  if (ans)
    cout << sum << endl;
  else
    cout << "-1" << endl;
}
