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 double Eps = 1e-8;
const long long Inf = 1e18 + 7;
const int Mod = 1e9 + 7;
const long long MAXN = 1e5 + 7;
vector<long long> edge[MAXN];
long long s[MAXN], a[MAXN];
long long dfs(long long i, long long val) {
  long long ret = 0, minn = Inf;
  if (s[i] == -1) {
    for (auto v : edge[i]) {
      if (s[v] - val < 0) {
        return -1;
      }
      if (minn > s[v] - val) {
        minn = s[v] - val;
      }
    }
    if (minn != Inf)
      a[i] = minn;
    else
      a[i] = 0;
    s[i] = val + a[i];
  } else {
    a[i] = s[i] - val;
  }
  for (auto v : edge[i]) {
    long long tmp = dfs(v, s[i]);
    if (tmp != -1) {
      ret += tmp;
    } else {
      return -1;
    }
  }
  return ret + a[i];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout << setprecision(numeric_limits<double>::max_digits10 + 1);
  long long n;
  cin >> n;
  a[1] = 1;
  for (int i = 2; i <= n; i++) {
    long long tmp;
    cin >> tmp;
    edge[tmp].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
  }
  cout << dfs(1, 0) << '\n';
}
