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 int N = 1000 * 100 + 20;
const long long INF = (long long)1e18;
long long n, ans;
long long a[N], s[N], p[N];
vector<long long> adj[N];
bool vis[N];
void dfs(int v, int sum) {
  vis[v] = true;
  if (s[v] != -1) a[v] = s[v] - sum;
  if (s[v] == -1) {
    long long mini = INF;
    for (auto u : adj[v])
      if (!vis[u]) mini = min(mini, s[u]);
    a[v] = (mini == INF ? 0 : mini - sum);
  }
  for (auto u : adj[v])
    if (!vis[u]) dfs(u, sum + a[v]);
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  cin >> n;
  for (int i = 2; i <= n; i++) {
    cin >> p[i];
    adj[p[i]].push_back(i);
    adj[i].push_back(p[i]);
  }
  for (int i = 1; i <= n; i++) cin >> s[i];
  dfs(1, 0);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] < 0) return cout << -1 << endl, 0;
    ans += a[i];
  }
  cout << ans << endl;
  return 0;
}
