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 mod = (1e+9) + 7;
const long long size = 2e5 + 9;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long power(long long a, long long b, long long m) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % m;
    b = b >> 1;
    a = (a * a) % m;
  }
  return res;
}
long long par[size];
long long s[size];
long long a[size];
vector<long long> chi[size];
void give(long long x) {
  if (x == 1) {
  }
  for (long long i = 0; i < chi[x].size(); i++) {
    if (s[chi[x][i]] != -1) {
      a[chi[x][i]] = s[chi[x][i]] - s[x];
      give(chi[x][i]);
    } else {
      long long cc = chi[x][i];
      if (chi[cc].size() == 0)
        s[cc] = s[x];
      else
        s[cc] = s[chi[cc][0]];
      for (long long j = 1; j < chi[cc].size(); j++) {
        s[cc] = min(s[cc], s[chi[cc][j]]);
      }
      a[cc] = s[cc] - s[x];
      give(chi[x][i]);
    }
  }
}
void solve() {
  long long n;
  cin >> n;
  for (long long i = 1; i < n; i++) {
    cin >> par[i + 1];
    chi[par[i + 1]].push_back(i + 1);
  }
  for (long long i = 1; i < n + 1; i++) cin >> s[i];
  a[1] = s[1];
  give(1);
  long long sum = 0;
  for (long long i = 1; i < n + 1; i++) {
    if (a[i] < 0) {
      cout << "-1" << endl;
      return;
    }
    sum += a[i];
  }
  cout << sum << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  clock_t beg = clock();
  solve();
  clock_t end = clock();
  fprintf(stderr, "%.3f sec, Copyright %c 2019 Skyscraper. \n",
          double(end - beg) / CLOCKS_PER_SEC, 184);
  return 0;
}
