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 n;
long long ans = 0;
vector<long long> adj[(long long)1e5 + 1];
int p[(long long)1e5 + 1];
long long s[(long long)1e5 + 1], a[(long long)1e5 + 1];
bool f = 0;
void dfs(long long node, long long h) {
  if (h % 2 == 1) {
    for (long long i : adj[node]) {
      dfs(i, h + 1);
    }
    return;
  }
  if (adj[node].empty()) return;
  long long mini = 1e9 + 5;
  for (long long i : adj[node]) {
    if (s[i] < mini) mini = s[i];
  }
  a[node] = mini - s[p[node]];
  if (a[node] < 0) {
    f = 1;
  }
  for (long long i : adj[node]) {
    a[i] = s[i] - a[node] - s[p[node]];
    dfs(i, h + 1);
  }
  return;
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long temp;
  cin >> n;
  for (__typeof(n + 1) i = (2) - ((2) > (n + 1));
       i != (n + 1) - ((2) > (n + 1)); i += 1 - 2 * ((2) > (n + 1))) {
    cin >> p[i];
    adj[p[i]].emplace_back(i);
  }
  for (__typeof(n + 1) i = (1) - ((1) > (n + 1));
       i != (n + 1) - ((1) > (n + 1)); i += 1 - 2 * ((1) > (n + 1)))
    cin >> s[i];
  a[1] = s[1];
  dfs(1, 1);
  for (__typeof(n + 1) i = (1) - ((1) > (n + 1));
       i != (n + 1) - ((1) > (n + 1)); i += 1 - 2 * ((1) > (n + 1))) {
    ans += a[i];
  }
  if (f == 0)
    cout << ans;
  else
    cout << "-1";
}
