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 = 1e5 + 5;
vector<int> e[N];
int a[N], par[N], s[N];
long long sum;
void dfs(int x, int pa, bool odd = true) {
  if (odd && pa) a[x] = (s[x] < s[par[pa]]) ? -1 : (s[x] - s[par[pa]]);
  int minn = INT_MAX;
  for (int y : e[x]) {
    dfs(y, x, !odd);
    minn = min(minn, a[y]);
  }
  if (!odd && e[x].size()) {
    for (int y : e[x]) a[y] -= minn;
    a[x] = minn;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 2; i <= n; ++i) {
    int x;
    cin >> x;
    e[x].push_back(i);
    par[i] = x;
  }
  for (int i = 1; i <= n; ++i) cin >> s[i];
  a[1] = s[1];
  dfs(1, 0);
  for (int i = 1; i <= n; ++i) {
    if (a[i] < 0) return cout << -1 << endl, 0;
    sum += a[i];
  }
  cout << sum << endl;
  return 0;
}
