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<long long int> graph[100005];
long long int sum[100005];
long long int a[100005];
bool good = true;
void dfs(long long int n, long long int p) {
  long long int z = 10000000000000;
  bool pr = false;
  if (sum[n] != -1) {
    a[n] = sum[n] - p;
    for (long long int i = 0; i < graph[n].size(); i++) {
      dfs(graph[n][i], sum[n]);
    }
  } else {
    for (long long int i = 0; i < graph[n].size(); i++) {
      pr = true;
      z = min(z, sum[graph[n][i]]);
    }
    if (z < p) {
      good = false;
    }
    if (pr) {
      a[n] = z - p;
    } else {
      a[n] = 0;
    }
    for (long long int i = 0; i < graph[n].size(); i++) {
      dfs(graph[n][i], z);
    }
  }
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n;
  cin >> n;
  for (long long int i = 2; i <= n; i++) {
    long long int z;
    cin >> z;
    graph[z].push_back(i);
  }
  for (long long int i = 1; i <= n; i++) {
    cin >> sum[i];
  }
  dfs(1, 0);
  if (!good) {
    cout << -1;
    return 0;
  }
  long long int ans = 0;
  a[1] = sum[1];
  for (long long int i = 1; i <= n; i++) {
    ans += a[i];
  }
  cout << ans;
}
