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 father[100006], cost[100006];
vector<vector<long>> childs(100006, vector<long>());
bool sol = true;
long long totalSum = 0;
void dfs(long node, long depth) {
  long long minCost = 9999999999;
  if ((depth % 2) == 0) {
    cost[node] = cost[father[node]];
    if (childs[node].size() > 0) {
      for (auto child : childs[node]) {
        minCost = min(minCost, cost[child]);
      }
      cost[node] = minCost;
    }
  }
  totalSum += cost[node];
  if (node != 1) {
    totalSum -= cost[father[node]];
  }
  for (auto child : childs[node]) {
    dfs(child, depth + 1);
    if ((depth % 2) == 0) {
      if (cost[child] < cost[father[node]]) {
        sol = false;
      }
    }
  }
}
int main() {
  long n;
  cin >> n;
  for (int i = 2; i <= n; i++) {
    cin >> father[i];
    childs[father[i]].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    cin >> cost[i];
  }
  dfs(1, 1);
  if (sol) {
    cout << totalSum;
  } else {
    cout << "-1";
  }
}
