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 int M = 1e9 + 7;
const int N = 2e5;
long long po(int, int);
vector<int> e[N];
int par[N];
int dep[N];
long long s[N];
void dfs(int v) {
  if (dep[v] % 2 == 1) {
    long long mn = LLONG_MAX;
    long long mx = s[par[v]];
    for (auto i : e[v]) {
      mn = min(mn, s[i]);
    }
    if (mn < mx) {
      cout << "-1\n";
      exit(0);
    }
    if (mn == LLONG_MAX)
      s[v] = mx;
    else
      s[v] = mn;
  }
  for (auto i : e[v]) {
    dep[i] = dep[v] + 1;
    dfs(i);
  }
}
void solve() {
  int n;
  cin >> n;
  for (int i = 2; i < n + 1; i++) {
    int x;
    cin >> x;
    e[x].push_back(i);
    par[i] = x;
  }
  for (int i = 1; i < n + 1; i++) cin >> s[i];
  dfs(1);
  long long ans = 0;
  for (int i = 1; i < n + 1; i++) ans += s[i] - s[par[i]];
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
long long po(int a, int b) {
  if (b == 0) return 1;
  long long ans = 1;
  if (b % 2 == 0) {
    ans = po(a, b / 2) % M;
    ans = (ans % M * ans % M) % M;
  } else {
    ans = po(a, (b - 1) / 2) % M;
    ans = (ans % M * ans % M * a) % M;
  }
  return ans % M;
}
