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 n, m, k, x = 1, y, z, s[500005], p[500005], a[500005], ans, val;
vector<long long> v[500005];
bool fl;
void dfs(long long node) {
  if (x % 2 == 0) {
    val = 2e9, fl = 0;
    for (auto i : v[node]) {
      fl = 1;
      val = min(val, s[i]);
      if (s[i] < s[node]) cout << -1, exit(0);
    }
    if (fl) {
      a[node] = val - s[node];
      s[node] = val;
    }
  }
  x++;
  for (auto i : v[node]) {
    if (x % 2)
      a[i] = s[i] - s[node];
    else
      a[i] = 0;
    if (a[i] < 0 || (x % 2 && s[i] < 0)) cout << -1, exit(0);
    s[i] = s[node] + a[i];
    dfs(i);
  }
  x--;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (long long i = 2; i < n + 1; i++) cin >> p[i], v[p[i]].push_back(i);
  for (long long i = 1; i < n + 1; i++) cin >> s[i];
  a[1] = s[1];
  dfs(1);
  for (long long i = 1; i < n + 1; i++) ans += a[i];
  cout << ans;
  return 0;
}
