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 long M = 1e9 + 7;
int k;
int n;
vector<int> gr[100002];
int a[100002];
int b[100002];
int v[100002];
int mn;
bool ok = true;
void dfs(int x) {
  if (b[x] == -1) {
    mn = 1e9;
    for (auto y : gr[x]) {
      if (mn > b[y] - b[a[x]]) mn = b[y] - b[a[x]];
    }
    if (mn < 0) {
      ok = false;
      return;
    }
    if (gr[x].size() == 0) mn = 0;
    v[x] = mn;
    b[x] = v[x] + b[a[x]];
  } else {
    v[x] = b[x] - b[a[x]];
  }
  for (auto y : gr[x]) dfs(y);
}
int main() {
  cin >> n;
  a[1] = 0;
  for (int i = 2; i <= n; i++) {
    cin >> a[i];
    gr[a[i]].push_back(i);
  }
  for (int i = 0; i < n; i++) cin >> b[i + 1];
  dfs(1);
  if (ok) {
    long long sum = 0;
    for (int i = 1; i <= n; i++) sum += v[i];
    cout << sum;
  } else {
    cout << -1;
  }
}
