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 = 1e18;
const long long MOD = 1e9 + 7;
const long double eps = 1e-7;
const long long zero = 0;
vector<long long> s;
vector<long long> a;
vector<vector<long long>> g;
void update(long long cur, long long gone) {
  if (s[cur] != -1 && gone > s[cur]) {
    cout << -1 << endl;
    exit(0);
  }
  if (g[cur].size() == 0) {
    a[cur] = max(a[cur], zero);
    return;
  }
  for (long long i = 0; i < g[cur].size(); ++i)
    update(g[cur][i], max(gone, s[cur]));
  if (a[cur] == -1) {
    long long m = INF;
    for (long long i = 0; i < g[cur].size(); ++i)
      m = min(m, a[g[cur][i]] - gone);
    if (m < 0) {
      cout << -1 << endl;
      exit(0);
    }
    a[cur] = m;
    for (long long i = 0; i < g[cur].size(); ++i) a[g[cur][i]] -= m + gone;
  } else {
  }
}
int32_t main() {
  ios_base::sync_with_stdio(NULL);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(10);
  long long n;
  cin >> n;
  s.assign(n, 0);
  a.assign(n, 0);
  g.assign(n, vector<long long>(0));
  for (long long i = 1; i < n; ++i) {
    long long p;
    cin >> p;
    g[p - 1].push_back(i);
  }
  for (long long i = 0; i < n; ++i) cin >> s[i];
  a[0] = s[0];
  for (long long i = 1; i < n; ++i) a[i] = s[i];
  for (long long i = 0; i < g[0].size(); ++i) update(g[0][i], a[0]);
  long long res = 0;
  for (long long i = 0; i < a.size(); ++i) {
    res += a[i];
    if (a[i] < 0) {
      cout << -1 << endl;
      return 0;
    }
  }
  cout << res << endl;
  return 0;
}
