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 MAXn = 1e5 + 10;
vector<int> g[MAXn];
long long ar[MAXn], n, ans;
void dfs(int v, long long sum) {
  if (ar[v] != -1 && ar[v] < sum) {
    cout << -1;
    exit(0);
  }
  if (ar[v] == -1) {
    for (int i = 0; i < g[v].size(); i++)
      if (ar[v] == -1 || ar[g[v][i]] < ar[v]) ar[v] = ar[g[v][i]];
    if (ar[v] == -1) ar[v] = sum;
    if (ar[v] < sum) {
      cout << -1;
      exit(0);
    }
  }
  ans += ar[v] - sum;
  for (int i = 0; i < g[v].size(); i++) dfs(g[v][i], ar[v]);
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 2; i <= n; i++) {
    int p;
    cin >> p;
    g[p].push_back(i);
  }
  for (int i = 1; i <= n; i++) cin >> ar[i];
  dfs(1, 0);
  cout << ans;
}
