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 int MOD = 998244353;
long long int mod = 1e9 + 7;
long long int INF = 1e18;
long long int dx[] = {1, -1, 0, 0};
long long int dy[] = {0, 0, 1, -1};
long long int a[100009] = {0};
long long int sum[100009];
long long int par[100009];
bool dfs(vector<long long int> adj[], long long int x, long long int h,
         long long int s) {
  if (h % 2 == 1) {
    a[x] = sum[x] - s;
    s = sum[x];
  }
  if (a[x] < 0) return false;
  long long int m = INF;
  for (auto j : adj[x]) {
    if (j != par[x]) {
      par[j] = x;
      bool flag = dfs(adj, j, h + 1, s);
      if (!flag) return false;
      m = min(a[j], m);
    }
  }
  if (m != INF && h % 2 == 0) {
    a[x] += m;
    for (auto j : adj[x]) {
      if (j != par[x]) a[j] -= m;
    }
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n;
  cin >> n;
  vector<long long int> adj[n + 1];
  for (int i = 2; i <= n; i++) {
    long long int x;
    cin >> x;
    adj[x].push_back(i);
    adj[i].push_back(x);
  }
  for (int i = 1; i <= n; i++) cin >> sum[i];
  bool flag = dfs(adj, 1, 1, 0);
  if (!flag)
    cout << -1;
  else {
    long long int ans = 0;
    for (int i = 1; i <= n; i++) ans += a[i];
    cout << ans;
  }
  return 0;
}
