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;
int main() {
  long long int i, j, n, x, y, z, temp;
  cin >> n;
  long long a[n], b[n];
  vector<long long> v[100005];
  for (i = 1; i < n; i++) {
    cin >> a[i];
    v[a[i]].push_back(i + 1);
    v[i + 1].push_back(a[i]);
  }
  for (i = 0; i < n; i++) cin >> b[i];
  queue<long long> q;
  vector<long long> v1;
  q.push(1);
  long long vis[100005] = {0}, sum = 1, ans = b[0], f = 0;
  vis[1] = 1;
  while (!q.empty()) {
    x = q.front();
    q.pop();
    sum = 0;
    long long min = 1e9, cnt = 0;
    for (i = 0; i < v[x].size(); i++) {
      if (vis[v[x][i]] == 0) {
        q.push(v[x][i]);
        vis[v[x][i]] = 1;
        if (b[v[x][i] - 1] == -1) {
          b[v[x][i] - 1] = b[x - 1];
        } else {
          cnt++;
          sum += b[v[x][i] - 1] - b[x - 1];
          if ((b[v[x][i] - 1] - b[x - 1]) < min)
            min = b[v[x][i] - 1] - b[x - 1];
        }
      }
    }
    if (cnt >= 1) {
      sum -= min * (cnt - 1);
      ans += sum;
    }
    if (sum < 0 || min < 0) f = 1;
  }
  if (ans < b[0] || f == 1) {
    cout << "-1";
    return 0;
  }
  cout << ans << endl;
}
