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 mxN = 1e5 + 5;
vector<long long> adj[mxN];
vector<long long> min_adj(mxN, (long long)1e10 + 1);
vector<long long> parents(mxN, -1);
bool ok = 1;
vector<long long> ans, sums(mxN);
void dfs(long long S, long long cur_sum, long long prev) {
  if (S == prev) return;
  if (!ok) return;
  if (sums[S] == -1) {
    if (min_adj[S] != 1e10 + 1) {
      if (min_adj[S] < sums[parents[S]]) {
        ok = 0;
        return;
      }
      ans.push_back(min_adj[S] - sums[parents[S]]);
      sums[S] = min_adj[S];
    }
  } else {
    if (S != 0 && ok) {
      ans.push_back(sums[S] - sums[parents[S]]);
      cur_sum += sums[S] - sums[parents[S]];
    }
  }
  for (auto u : adj[S]) {
    dfs(u, cur_sum, S);
  }
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  long long N;
  cin >> N;
  for (long long i = 0; i < N - 1; i++) {
    long long A;
    cin >> A;
    A--;
    adj[A].push_back(i + 1);
    parents[i + 1] = A;
  }
  for (long long i = 0; i < N; i++) {
    cin >> sums[i];
  }
  for (long long i = 0; i < N; i++) {
    for (auto j : adj[i]) {
      min_adj[i] = min(min_adj[i], sums[j]);
    }
  }
  dfs(0, sums[0], -1);
  long long sum = sums[0];
  for (auto x : ans) sum += x;
  if (!ok)
    cout << -1 << "\n";
  else
    cout << sum << "\n";
}
