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 i, j, k, l, n, m, timer, res;
long long s[200009], h[200009], d[200009], num[200009];
vector<long long> a[200009];
long long dfs(long long v, long long parent, long long d, long long sum) {
  if (s[v] != -1 && sum > s[v]) {
    cout << -1 << '\n';
    exit(0);
  }
  if (s[v] != -1) num[v] = s[v] - sum;
  if (s[v] != -1) sum = s[v];
  long long k = ((long long)1e18 + 9ll);
  if (s[v] == -1) {
    for (auto to : a[v]) {
      if (to == parent) continue;
      k = min(k, s[to] - sum);
    }
  }
  if (k != ((long long)1e18 + 9ll)) sum += k, num[v] = k;
  if (k != ((long long)1e18 + 9ll) && k < 0) {
    cout << -1 << '\n';
    exit(0);
  }
  for (auto to : a[v]) {
    if (to != parent) {
      dfs(to, v, d + 1, sum);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (i = 2; i <= n; i++) {
    cin >> k;
    a[k].push_back(i);
    a[i].push_back(k);
  }
  for (i = 1; i <= n; i++) {
    cin >> s[i];
  }
  dfs(1, -1, 0, 0);
  for (i = 1; i <= n; i++) res += num[i];
  cout << res << '\n';
  exit(0);
}
