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 int power(long long int a, long long int b) {
  if (b == 0) return 1;
  if (b == 1) return a;
  long long int c = power(a, b / 2);
  if (b % 2 == 0) return ((c % 1000000007) * (c % 1000000007)) % 1000000007;
  return ((((c % 1000000007) * (c % 1000000007)) % 1000000007) *
          (a % 1000000007)) %
         1000000007;
}
vector<long long int> v[100005];
int vis[100005] = {0};
long long int a[100005], s[100005], n;
void dfs(long long int x, long long int sum) {
  if (s[x] != -1) {
    if (s[x] < sum) {
      cout << "-1\n";
      exit(0);
    }
  }
  if (v[x].size() == 0) {
    if (s[x] == -1) {
      a[x] = 0;
    }
  } else {
    long long int mn = 1000000007;
    for (auto z : v[x]) mn = min(mn, s[z]);
    if (mn == -1) mn = 0;
    if (s[x] == -1) {
      a[x] = mn;
      s[x] = mn;
    } else
      a[x] = mn + s[x];
    sum = max(sum, s[x]);
    for (auto z : v[x]) {
      dfs(z, sum);
      if (v[z].size() == 0 && s[z] == -1) {
        a[z] = 0;
        s[z] = s[x];
        continue;
      }
      a[z] = s[z] - a[x];
      if (a[z] < 0) a[z] = 0;
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int k, t, c = 0, i, j, m, cs = 0;
  t = 1;
  while (t--) {
    cin >> n;
    for (i = 1; i < n; i++) {
      cin >> m;
      v[m - 1].push_back(i);
    }
    for (i = 0; i < n; i++) {
      cin >> s[i];
    }
    m = s[0];
    dfs(0, 0);
    if (a[0] != m || s[0] != m)
      cout << "-1" << a[0];
    else {
      for (i = 0; i < n; i++) {
        c += a[i];
      }
      cout << c;
    }
    if (t != 0) cout << "\n";
  }
  return 0;
}
