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> v[100005];
int a[100005], s[100005];
bool vis[100005];
bool correct = true;
void dfs(int u, int sum) {
  vis[u] = true;
  if (s[u] == -1) {
    int mini = INT_MAX;
    for (auto it2 : v[u]) {
      mini = min(mini, s[it2]);
    }
    if (mini == INT_MAX) {
      a[u] = 0;
    } else {
      if (mini >= sum) {
        a[u] = mini - sum;
        for (auto it2 : v[u]) a[it2] = s[it2] - mini;
      } else {
        correct = false;
      }
    }
    if (correct == false) return;
    for (auto it : v[u]) {
      for (auto it2 : v[it]) {
        dfs(it2, s[it]);
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, x;
  cin >> n;
  for (int i = 2; i <= n; i++) {
    cin >> x;
    v[x].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
  }
  memset(a, -1, sizeof(a));
  memset(vis, false, sizeof(vis));
  a[1] = s[1];
  for (auto it : v[1]) {
    dfs(it, s[1]);
  }
  if (correct == true) {
    long long int sum = 0;
    for (int i = 1; i <= n; i++) sum += a[i];
    cout << sum << endl;
  } else {
    cout << "-1" << endl;
  }
  return 0;
}
