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;
void fun() {}
int flag = 0;
void dfs(int par, int pare, int vis[], vector<int> g[], long long int sumarr[],
         long long int a[]) {
  if (sumarr[par] == -1) {
    long long int minval = -1;
    for (int i = 0; i < g[par].size(); i++) {
      int child = g[par][i];
      if (!vis[child]) {
        if (minval == -1) minval = sumarr[child];
        minval = min(minval, sumarr[child]);
        vis[child] = 1;
        dfs(child, par, vis, g, sumarr, a);
      }
    }
    a[par] = minval - sumarr[pare];
    if (minval == -1) {
      a[par] = 0;
    }
    if (minval != -1 && a[par] < 0) {
      flag = 1;
      return;
    }
    for (int i = 0; i < g[par].size(); i++) {
      int child = g[par][i];
      a[child] = sumarr[child] - minval;
    }
  } else {
    for (int i = 0; i < g[par].size(); i++) {
      int child = g[par][i];
      if (!vis[child]) {
        vis[child] = 1;
        dfs(child, par, vis, g, sumarr, a);
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  fun();
  int n;
  cin >> n;
  long long int par[n + 1];
  par[1] = 0;
  par[0] = 0;
  for (int i = 2; i <= n; i++) {
    cin >> par[i];
  }
  long long int s[n + 1];
  s[0] = 0;
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
    if (s[i] == -1) {
      s[i] = LLONG_MAX;
    }
  }
  for (int i = 2; i <= n; i++) {
    s[par[i]] = min(s[par[i]], s[i]);
  }
  long long int ans = 0;
  for (int i = 1; i <= n; i++) {
    if (s[i] < s[par[i]]) {
      cout << "-1\n";
      return 0;
    }
    if (s[i] == LLONG_MAX) {
      s[i] = s[par[i]];
    }
    ans += s[i] - s[par[i]];
  }
  cout << ans << "\n";
  return 0;
}
