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> adj[100005];
long long par[100005], s[100005], a[100005], xm[100005];
void dfs(long long node) {
  if (node == 1)
    a[node] = s[node];
  else {
    if (s[node] == -1) {
      if ((long long int)adj[node].size() == 1)
        xm[node] = s[par[node]];
      else
        for (auto i : adj[node])
          if (i != par[node])
            xm[node] = (xm[node] == -1) ? s[i] : min(s[i], xm[node]);
      a[node] = xm[node] - s[par[node]];
      s[node] = s[par[node]] + a[node];
    } else {
      a[node] = s[node] - s[par[node]];
    }
  }
  for (auto i : adj[node])
    if (i != par[node]) dfs(i);
}
bool dfs1(long long node, long long mx) {
  bool f = 1;
  if (s[node] != -1) f = (s[node] >= mx);
  for (auto i : adj[node])
    if (i != par[node]) {
      if (s[node] != -1)
        f &= dfs1(i, s[node]);
      else
        f &= dfs1(i, mx);
    }
  return f;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long TESTS = 1;
  while (TESTS--) {
    long long n, i, j, k;
    cin >> n;
    for (i = 2; i <= n; i++) {
      cin >> par[i];
      adj[i].push_back(par[i]);
      adj[par[i]].push_back(i);
    }
    for (i = 0; i < 100005; i++) xm[i] = -1;
    for (i = 1; i <= n; i++) cin >> s[i];
    if (!dfs1(1, s[1])) {
      cout << "-1";
      return 0;
    }
    dfs(1);
    long long sum = 0;
    for (i = 1; i <= n; i++) sum += a[i];
    cout << sum;
  }
  return 0;
}
