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;
const long long mod = 1e9 + 7;
long double PI = acosl(-1);
const long long inf = 1e15;
const int nmax = 1e5 + 5;
vector<int> g[nmax];
bool posibul = true;
long long s[nmax];
long long a[nmax];
long long ans = 0;
void dfs(int u, int lvl, int par) {
  if (lvl & 1) {
    if (a[par] > s[u])
      posibul = false;
    else {
      ans += s[u] - a[par];
      a[u] = s[u];
    }
  }
  vector<pair<int, int> > tmp;
  if (lvl % 2 == 0) {
    for (auto v : g[u]) {
      tmp.push_back({s[v], v});
    }
  }
  sort(tmp.begin(), tmp.end());
  if (lvl % 2 == 0) {
    if (tmp.size()) {
      int v = tmp[0].second;
      long long ss = tmp[0].first;
      a[u] = a[par];
      if (ss > a[u]) ans += (ss - a[u]), a[u] += (ss - a[u]);
    }
  }
  for (auto v : g[u]) {
    dfs(v, lvl + 1, u);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 2; i <= n; i++) {
    int u;
    cin >> u;
    g[u].push_back(i);
  }
  for (int i = 1; i <= n; i++) cin >> s[i];
  a[1] = s[1];
  dfs(1, 1, 0);
  if (posibul)
    cout << ans;
  else
    cout << "-1";
}
