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 a[100001], val[100001];
vector<int> g[100001];
long long int n;
void dfs(int u, long long int s) {
  if (a[u] == -1) {
    a[u] = s;
    long long int cur = 1LL * 1e18;
    for (int i = 0; i < g[u].size(); i++) {
      int v = g[u][i];
      cur = min(cur, a[v]);
    }
    if (cur != 1LL * 1e18) a[u] = cur;
  }
  if (a[u] != -1) {
    if (s > a[u]) {
      cout << -1 << endl;
      exit(0);
    }
    val[u] = a[u] - s;
    s = a[u];
    for (int i = 0; i < g[u].size(); i++) {
      int v = g[u][i];
      dfs(v, s);
    }
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int p;
    cin >> p;
    g[p].push_back(i + 2);
  }
  for (int i = 1; i <= n; i++) cin >> a[i];
  dfs(1, 0);
  long long int ans = 0;
  for (int i = 0; i < n + 1; i++) ans += val[i];
  cout << ans << endl;
  return 0;
}
