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 int N = 5e5 + 5;
const int MOD = 1e9 + 7;
vector<int> g[N];
long long s[N];
long long a[N];
long long ans = 0;
long long sz[N];
long long sum[N];
long long ret[N];
void dfs(int v, int p) {
  if (s[v] == -1)
    sum[v] = sum[p];
  else
    sum[v] = s[v];
  a[v] = LLONG_MAX;
  for (auto &it : g[v]) {
    if (it != p) {
      dfs(it, v);
      a[v] = min(a[it], a[v]);
      ret[v] += ret[it];
    }
  }
  if (a[v] == LLONG_MAX) a[v] = 0;
  if (s[v] == -1) {
    ret[v] -= (g[v].size()) * a[v];
  } else {
    a[v] = sum[v] - sum[p];
    if (a[v] < 0) {
      cout << -1;
      exit(0);
    }
  }
  ret[v] += a[v];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int p;
    cin >> p;
    --p;
    g[p].push_back(i);
  }
  for (int i = 0; i < n; ++i) {
    cin >> s[i];
  }
  dfs(0, 0);
  cout << ret[0] + s[0];
  return 0;
}
