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;
using ll = long long;
ll mod = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ll t, n, m, i, j, k;
  cin >> n;
  vector<ll> v[100001];
  ll a[100001] = {0};
  ll c[n + 1], b[n + 1];
  for (i = 2; i <= n; i++) {
    cin >> k;
    b[i] = k;
    v[k].push_back(i);
  }
  for (i = 1; i <= n; i++) cin >> c[i];
  for (i = 1; i <= n; i++) {
    if (c[i] == -1) {
      k = INT_MAX;
      for (j = 0; j < v[i].size(); j++) {
        k = min(k, c[v[i][j]]);
      }
      if (k == INT_MAX) k = c[b[i]];
      a[i] = k;
    } else
      a[i] = c[i];
  }
  ll ans = c[1];
  for (i = 1; i <= n; i++) {
    for (j = 0; j < v[i].size(); j++) {
      ans += a[v[i][j]] - a[i];
      if (a[v[i][j]] - a[i] < 0) {
        ans = -1;
        break;
      }
    }
    if (ans == -1) break;
  }
  cout << ans;
}
