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 s[100100], arr[100100], lvl[100100], par[100100];
vector<int> adj[100100];
void dfs(int src, int p, int a, int l) {
  lvl[src] = l;
  l++;
  par[src] = p;
  if (src > 1) arr[src] = max(0, s[src] - a);
  if (s[src] != -1 && s[src] < a) {
    printf("%s", "-1\n");
    exit(0);
  }
  for (int e : adj[src]) {
    if (e == p) continue;
    if (s[src] == -1)
      dfs(e, src, a, l);
    else
      dfs(e, src, s[src], l);
  }
}
bool cmp(int a, int b) { return lvl[a] > lvl[b]; }
int main() {
  int i, j, k;
  int n, m;
  scanf("%d", &n);
  for (i = 2; i <= n; i++) {
    scanf("%d", &k);
    adj[i].push_back(k);
    adj[k].push_back(i);
  }
  for (i = 1; i <= n; i++) scanf("%d", &s[i]);
  memset(arr, 0, sizeof(arr));
  arr[1] = s[1];
  dfs(1, -1, s[1], 1);
  vector<int> vec;
  for (i = 1; i <= n; i++) {
    if (s[i] == -1) vec.push_back(i);
  }
  sort(vec.begin(), vec.end(), cmp);
  int mn;
  for (int e : vec) {
    mn = 1000000000;
    for (int x : adj[e]) {
      if (x != par[e]) mn = min(mn, arr[x]);
    }
    if (mn == 1000000000) continue;
    arr[e] = mn;
    for (int x : adj[e]) {
      if (x != par[e]) arr[x] -= arr[e];
    }
  }
  long long int cnt = 0;
  for (i = 1; i <= n; i++) cnt += arr[i];
  cout << cnt << endl;
  return 0;
}
