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;
long long s[100010];
long long a[100010];
vector<int> tree[100010];
int n;
void dfs(int x, int p, int d) {
  if (d % 2 == 0) {
    long long su = LLONG_MAX;
    for (auto y : tree[x]) {
      su = min(su, s[y]);
    }
    if (su == LLONG_MAX) {
      a[x] = 0;
    } else {
      s[x] = su;
      a[x] = s[x] - s[p];
    }
  } else {
    if (x != 0) a[x] = s[x] - s[p];
  }
  for (auto y : tree[x]) {
    dfs(y, x, d + 1);
  }
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int x;
    cin >> x;
    tree[x - 1].push_back(i);
  }
  for (int i = 0; i < n; ++i) {
    cin >> s[i];
  }
  a[0] = s[0];
  dfs(0, 0, 1);
  long long sum = 0;
  bool puedo = true;
  for (int i = 0; i < n; ++i) {
    sum += a[i];
    if (a[i] < 0) {
      puedo = false;
    }
  }
  if (!puedo) {
    cout << -1 << '\n';
  } else {
    cout << sum << '\n';
  }
  return 0;
}
