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;
int main() {
  int n, x;
  bool state = true;
  scanf("%d", &n);
  int sum[n + 1], arr[n + 1];
  vector<int> graph[n + 1];
  for (int i = 2; i <= n; i++) {
    scanf("%d", &x);
    graph[x].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &sum[i]);
  }
  arr[1] = sum[1];
  queue<pair<int, int> > q;
  q.push(make_pair(1, sum[1]));
  while (!q.empty() && state) {
    pair<int, int> p = q.front();
    q.pop();
    int sz = graph[p.first].size();
    if (sz == 0) {
      if (sum[p.first] == -1) arr[p.first] = 0;
      continue;
    }
    int M = 1000000000;
    for (int i = 0; i < sz; i++) {
      int v = graph[p.first][i];
      M = min(sum[v], M);
      if (sum[v] != -1 && sum[v] < p.second) {
        state = false;
      }
    }
    if (sum[graph[p.first][0]] != -1) arr[p.first] = M - p.second;
    for (int i = 0; i < sz; i++) {
      int v = graph[p.first][i];
      if (sum[v] != -1) {
        arr[v] = sum[v] - M;
        q.push(make_pair(v, sum[v]));
      } else {
        q.push(make_pair(v, p.second));
      }
    }
  }
  long long ans = 0;
  if (state) {
    for (int i = 1; i <= n; i++) {
      ans += (1ll * arr[i]);
    }
    cout << ans << endl;
  } else
    printf("-1\n");
}
