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 inf = (long long)9e18;
struct cmp {
  bool operator()(const long long &i, const long long &j) const {
    return i > j;
  }
};
long long ans = 0;
long long sumt = 0;
long long sumd = 0;
vector<vector<long long>> g;
vector<long long> data1;
long long dfs(long long v, long long pp) {
  if (data1[v] != -1) {
    long long e;
    if (data1[v] < pp) {
      cout << -1;
      exit(0);
    }
    long long t = data1[v] - pp;
    for (long long i = 0; i < g[v].size(); i++) {
      long long to = g[v][i];
      long long zz = dfs(to, data1[v]);
    }
    if (t < 0) {
      cout << -1;
      exit(0);
    }
    return t;
  } else {
    if (g[v].size() == 0) return 0;
    vector<long long> A(g[v].size(), 0);
    for (long long i = 0; i < g[v].size(); i++) {
      long long to = g[v][i];
      A[i] = dfs(to, pp);
    }
    long long mi = inf;
    for (long long i = 0; i < A.size(); i++) {
      if (A[i] < mi) mi = A[i];
    }
    for (long long i = 0; i < A.size(); i++) {
      ans += A[i] - mi;
    }
    ans += mi;
    return 0;
  }
}
void dfs1(long long v) {}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long sum = 0;
  long long n;
  cin >> n;
  data1.resize(n + 1);
  g.resize(n + 1);
  vector<long long> pr(n + 1);
  for (long long i = 2; i <= n; i++) {
    cin >> pr[i];
    g[pr[i]].push_back(i);
  }
  for (long long i = 1; i <= n; i++) {
    cin >> data1[i];
  }
  sumt = data1[1];
  long long awd = dfs(1, data1[1]);
  ans += data1[1];
  cout << ans;
  return 0;
}
