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<int> tree[200005];
long long int second[200005];
long long int a[200005];
bool flag = true;
void dfs(int root, int par) {
  if (second[root] == -1) {
    long long int mi = INT_MAX;
    int cnt = 0;
    for (auto i : tree[root]) {
      ++cnt;
      if (i != par) mi = min(mi, second[i]);
    }
    if (cnt == 0) {
      a[root] = 0;
    } else if (mi < second[par]) {
      flag = false;
    } else {
      a[root] = mi - second[par];
      second[root] = second[par] + a[root];
      for (auto i : tree[root]) {
        a[i] = second[i] - second[root];
        dfs(i, root);
      }
    }
  } else {
    for (auto i : tree[root]) {
      dfs(i, root);
    }
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 2; i <= n; ++i) {
    int x;
    cin >> x;
    tree[x].push_back(i);
  }
  for (int i = 1; i <= n; ++i) {
    cin >> second[i];
  }
  a[1] = second[1];
  dfs(1, 0);
  long long int ans = 0;
  for (int i = 1; i <= n; ++i) ans += a[i];
  if (flag)
    cout << ans;
  else
    cout << -1;
  return 0;
}
