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 N = 1e5 + 2;
const long long mod = 1000000007;
long long n, par[N], s[N], a[N], d[N];
vector<long long> adj[N], dep[N];
void dfs(long long u, long long p) {
  for (long long v : adj[u]) {
    if (v == p) continue;
    d[v] = d[u] + 1;
    dep[d[v]].push_back(v);
    dfs(v, u);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cin >> n;
  for (int i = 2; i <= n; i++) {
    cin >> par[i];
    adj[i].push_back(par[i]);
    adj[par[i]].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
  }
  a[1] = s[1];
  dep[1].push_back(1), d[1] = 1;
  dfs(1, 1);
  for (long long dd = 2; dd <= n; dd += 2) {
    for (long long u : dep[dd]) {
      if (adj[u].size() == 1)
        a[u] = 0;
      else {
        long long mn = (1LL << 61);
        for (long long v : adj[u]) {
          if (v == par[u]) continue;
          mn = min(mn, s[v] - s[par[u]]);
        }
        if (mn < 0) {
          cout << -1;
          return 0;
        }
        a[u] = mn;
        for (long long v : adj[u]) {
          if (v == par[u]) continue;
          a[v] = s[v] - s[par[u]] - a[u];
        }
      }
    }
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += a[i];
  }
  cout << ans;
}
