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 + 50;
int n, m, u, v, k, cnt, cnt2;
vector<pair<int, int> > g[N];
bool visited[N];
int mx = INT_MIN;
const int fx[9] = {0, 0, 1, -1, 0, 1, 1, -1, -1};
const int fy[9] = {1, -1, 0, 0, 0, 1, -1, 1, -1};
const int dx[] = {0, 1, 0, -1};
const int dy[] = {1, 0, -1, 0};
int p[N];
long long second[N];
int main() {
  cin >> n;
  for (int i = 2; i <= n; i++) {
    cin >> p[i];
  }
  for (int i = 1; i <= n; i++) {
    cin >> second[i];
  }
  for (int i = 1; i <= n; i++) {
    if (second[i] == -1) {
      second[i] = 1e9 + 1;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (second[i] < 1e9 + 1) {
      second[p[i]] = min(second[i], second[p[i]]);
    }
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    if (second[i] < second[p[i]]) {
      cout << -1;
      return 0;
    }
    if (second[i] == 1e9 + 1) {
      continue;
    }
    ans += second[i] - second[p[i]];
  }
  cout << ans;
}
