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 INF = 1e9;
bool ans = true;
int a[100100];
int s[100100];
vector<int> edges[100100];
bool used[100100];
void dfs(int vert, int currSum) {
  used[vert] = true;
  if (s[vert] == -1) {
    int mn = INF;
    s[vert] = currSum;
    for (auto edge : edges[vert]) {
      if (!used[edge]) {
        mn = min(mn, s[edge]);
      }
    }
    if (mn != INF && mn > currSum) {
      s[vert] = mn;
    }
  }
  if (s[vert] < currSum) {
    ans = false;
    return;
  }
  a[vert] = s[vert] - currSum;
  for (auto i : edges[vert]) {
    if (!used[i]) dfs(i, s[vert]);
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 2; i <= n; i++) {
    int x;
    cin >> x;
    edges[x].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
  }
  dfs(1, 0);
  if (!ans) {
    cout << -1;
    return 0;
  }
  long long sum = 0;
  for (int i = 1; i <= n; i++) {
    sum += a[i];
  }
  cout << sum;
}
