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;
long mm = 1000000007;
const int maxn = 200005;
template <class TYPE>
TYPE gcd(TYPE x, TYPE y) {
  if (y)
    return gcd(y, x % y);
  else
    return x;
}
template <class TYPE>
TYPE lcm(TYPE x, TYPE y) {
  TYPE t = gcd(x, y);
  return t ? (x / t * y) : 0;
}
int n;
vector<int> tree[maxn];
int s[maxn];
long long ans;
bool check = true;
void dfs(int i, int pre) {
  if (s[i] == -1) {
    int Min = mm;
    for (int kk = 0; kk < tree[i].size(); kk++) Min = min(Min, s[tree[i][kk]]);
    if (Min == mm)
      s[i] = s[pre];
    else {
      s[i] = Min;
      ans += s[i] - s[pre];
    }
  } else {
    ans += s[i] - s[pre];
  }
  if (s[i] - s[pre] < 0) {
    check = false;
    return;
  }
  for (int kk = 0; kk < tree[i].size(); kk++) dfs(tree[i][kk], i);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int x;
    cin >> x;
    tree[x].push_back(i + 2);
  }
  for (int i = 1; i <= n; i++) cin >> s[i];
  s[0] = 0;
  dfs(1, 0);
  if (!check) ans = -1;
  cout << ans;
  return 0;
}
