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;
const long long int N = 1e5 + 2;
vector<long long int> g[N];
long long int sm[N], dep[N], a[N], second[N];
void dfs(long long int v, long long int p) {
  dep[v] = 1 + dep[p];
  long long int mn = 1e18;
  for (long long int u : g[v])
    if (u != p) {
      dfs(u, v);
      mn = min(mn, sm[u]);
    }
  if (mn == 1e18) return;
  if (dep[v] % 2 == 0) a[v] = mn - sm[p];
}
void dfs2(long long int v, long long int p) {
  if (dep[v] % 2)
    a[v] = sm[v] - sm[p];
  else
    sm[v] = sm[p] + a[v];
  for (long long int u : g[v])
    if (u != p) dfs2(u, v);
}
void solve() {
  long long int n;
  cin >> n;
  for (long long int i = 2; i <= n; i++) {
    long long int x;
    cin >> x;
    g[x].push_back(i);
    g[i].push_back(x);
  }
  for (long long int i = 1; i <= n; i++) cin >> sm[i];
  dfs(1, 0);
  dfs2(1, 0);
  long long int ans = 0;
  for (long long int i = 1; i <= n; i++)
    if (a[i] < 0) {
      cout << -1;
      return;
    }
  for (long long int i = 1; i <= n; i++) ans += a[i];
  cout << ans;
}
int32_t main() {
  ios::sync_with_stdio(false);
  long long int t = 1;
  while (t--) {
    solve();
  }
}
