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 ans = 0, l;
long long s[100010], mn[100010];
vector<int> a[100010];
void dfs(int x, long long tmp) {
  if (mn[x] < tmp && s[x] != -1) {
    l = 0;
  }
  if (mn[x] != 1e9 + 1) ans += mn[x] - tmp;
  for (int i = 0; i < a[x].size(); ++i) {
    dfs(a[x][i], max(tmp, mn[x]));
  }
}
int dfs1(int x) {
  if (s[x] != -1) mn[x] = s[x];
  int tmp = 1e9 + 1, tmp1;
  for (int i = 0; i < a[x].size(); ++i) {
    tmp1 = dfs1(a[x][i]);
    tmp = min(tmp, tmp1);
  }
  if (s[x] == -1)
    mn[x] = tmp;
  else
    mn[x] = s[x];
  return mn[x];
}
int main() {
  int n;
  while (cin >> n) {
    int x;
    for (int i = 1; i < n; ++i) {
      scanf("%d", &x);
      a[x].push_back(i + 1);
    }
    for (int i = 1; i <= n; ++i) {
      scanf("%lld", s + i);
    }
    l = 1;
    ans = 0;
    dfs1(1);
    dfs(1, 0);
    if (l)
      cout << ans << endl;
    else
      cout << -1 << endl;
  }
  return 0;
}
