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> pax[1000000 + 5];
int d[1000000 + 5];
int vis[1000000 + 5], val[1000000 + 5];
int flag, sum;
long long total;
void dfs(int curr, int parent, int depth) {
  if (depth % 2 == 1) {
    if (d[parent] > d[curr]) {
      flag = 1;
      return;
    }
    val[curr] = d[curr] - d[parent];
  } else {
    int men = pow(10, 6);
    for (auto x : pax[curr]) {
      if (x != parent) {
        men = min(d[x], men);
      }
    }
    if (men == pow(10, 6)) {
      men = d[parent];
    }
    if (men < d[parent]) {
      flag = 1;
    }
    val[curr] = men - d[parent];
    d[curr] = d[parent] + val[curr];
  }
  for (auto x : pax[curr]) {
    if (x != parent) {
      dfs(x, curr, depth + 1);
    }
  }
}
int main() {
  int n, x, l;
  cin >> n;
  for (int i = 2; i <= n; i++) {
    cin >> x;
    pax[x].push_back(i);
    pax[i].push_back(x);
  }
  for (int i = 1; i <= n; i++) {
    cin >> l;
    d[i] = l;
  }
  d[0] = 0;
  dfs(1, 0, 1);
  if (flag == 1) {
    cout << -1 << endl;
  } else {
    for (int i = 1; i <= n; i++) {
      total += val[i];
    }
    cout << total << endl;
  }
  return 0;
}
