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 long long int MAX = 1e5 + 2;
vector<long long int> adj[MAX];
vector<long long int> sum(MAX);
vector<long long int> A(MAX);
void dfs(long long int root, long long int prev, long long int level) {
  if (level & 1) {
    A[root] = sum[root] - sum[prev];
  } else {
    if (adj[root].size() == 1) {
      A[root] = 0;
      sum[root] = sum[prev];
      return;
    }
    long long int min1 = 1e18;
    for (long long int y : adj[root]) {
      if (y == prev) continue;
      min1 = min(min1, sum[y]);
    }
    A[root] = min1 - sum[prev];
    sum[root] = sum[prev] + A[root];
  }
  for (long long int y : adj[root]) {
    if (y == prev) continue;
    dfs(y, root, level + 1);
  }
}
int32_t main() {
  long long int n, p, ans = 0;
  cin >> n;
  for (long long int i = 2; i <= n; i++) {
    cin >> p;
    adj[p].push_back(i);
    adj[i].push_back(p);
  }
  for (long long int i = 1; i <= n; i++) {
    cin >> sum[i];
  }
  dfs(1, 0, 1);
  for (long long int i = 1; i <= n; i++) {
    if (A[i] < 0) {
      cout << -1 << '\n';
      return 0;
    }
    ans += A[i];
  }
  cout << ans << '\n';
}
