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<long long> adj[100050];
long long s[100050], a[100050], p[100050];
void dfs(long long u, long long depth = 1) {
  long long mn = 1e18, val = s[p[u]];
  if (!depth) {
    for (auto v : adj[u]) {
      if (s[v] < val) {
        cout << -1;
        exit(0);
      }
      mn = min(s[v] - val, mn);
    }
    a[u] = (mn == 1e18 ? 0 : mn);
    for (auto v : adj[u]) a[v] = s[v] - val - mn;
  }
  for (auto v : adj[u]) dfs(v, 1 - depth);
}
signed main() {
  long long n;
  cin >> n;
  for (long long i = 2; i <= n; i++) {
    long long q;
    cin >> q;
    adj[q].push_back(i);
    p[i] = q;
  }
  for (long long i = 1; i <= n; i++) cin >> s[i];
  dfs(1);
  long long ans = s[1];
  for (long long i = 2; i <= n; i++) ans += a[i];
  cout << ans;
}
