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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
bool flag;
long long int par[200005];
long long int val[200005];
long long int av[2000005];
vector<vector<long long int> > t;
void dfs(long long int r, long long int p, long long int n, long long int h) {
  if (h % 2 != 0) {
    if (val[r] >= n)
      av[r] = val[r] - n;
    else
      flag = false;
  } else {
    long long int m = 0x3f3f3f3f;
    for (auto i : t[r])
      if (i != p)
        m = min((long long int)m, (long long int)val[i] - n), av[r] = m;
    if (m < 0) flag = false;
  }
  for (auto i : t[r])
    if (i != p) dfs(i, r, n + av[r], h + 1);
}
signed main() {
  std::ios::sync_with_stdio(false);
  long long int n;
  cin >> n;
  t.clear();
  t.resize(n + 1);
  for (long long int i = 2; i < n + 1; i++)
    cin >> par[i], t[par[i]].push_back(i);
  for (long long int i = 1; i < n + 1; i++) cin >> val[i];
  flag = true;
  dfs(1, 0, 0, 1);
  long long int sum = 0;
  if (flag) {
    for (long long int i = 1; i < n + 1; i++) sum += av[i];
    cout << sum << "\n";
  } else
    cout << -1 << "\n";
}
