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;
int n;
vector<int> adj[100100];
long long Aval[100100], Sval[100100], sum = 0;
bool fail = false;
long long DFS(int here, int depth, long long curSum) {
  long long mn = (long long)1e15;
  for (int there : adj[here]) {
    mn = min(mn, DFS(there, depth + 1, curSum + Aval[here]));
  }
  if (depth % 2 == 1) {
    if (Sval[here] > mn) fail = true;
  } else {
    Sval[here] = mn;
  }
  return Sval[here];
}
void DFS2(int here, int depth, long long curSum) {
  if (Sval[here] == (long long)1e15) return;
  sum += Sval[here] - curSum;
  for (int there : adj[here]) {
    DFS2(there, depth + 1, Sval[here]);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    int p;
    scanf("%d", &p);
    adj[p].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &Sval[i]);
  }
  DFS(1, 1, 0);
  DFS2(1, 1, 0);
  if (!fail)
    printf("%lld", sum);
  else
    printf("-1");
}
