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;
template <class T>
ostream& operator<<(ostream& stream, const vector<T> v) {
  stream << "[ ";
  for (int i = 0; i < (int)v.size(); i++) stream << v[i] << " ";
  stream << "]";
  return stream;
}
long long fpow(long long x, long long p, long long m) {
  long long r = 1;
  for (; p; p >>= 1) {
    if (p & 1) r = r * x % m;
    x = x * x % m;
  }
  return r;
}
long long inv(long long a, long long b) {
  return a < 2 ? a : ((a - inv(b % a, a)) * b + 1) / a % b;
}
int gcd(int a, int b) {
  if (!b) return a;
  return gcd(b, a % b);
}
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
long long N, s[100010];
vector<int> g[100010];
long long res = 0;
bool ok = 1;
void dfs(int nod, int h, long long v) {
  if (h % 2) {
    if (s[nod] < v) ok = 0;
    res += s[nod] - v;
    for (int x : g[nod]) dfs(x, h + 1, s[nod]);
  } else {
    if (g[nod].size() == 0) return;
    long long minv = 1ll << 60;
    for (int x : g[nod]) minv = min(minv, s[x]);
    if (minv < v) ok = 0;
    res += minv - v;
    for (int x : g[nod]) dfs(x, h + 1, minv);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> N;
  int i;
  for (i = 2; i <= N; i++) {
    int p;
    cin >> p;
    g[p].push_back(i);
  }
  for (i = 1; i <= N; i++) cin >> s[i];
  dfs(1, 1, 0);
  if (!ok)
    cout << -1 << "\n";
  else
    cout << res << "\n";
}
