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 INF = 2147483647;
const long long INFL = 9223372036854775807LL;
const double EPSILON = 0.00000001;
const long long MOD = 1000000007;
long long parent[200000 + 5];
vector<long long> children[200000 + 5];
long long sum[200000 + 5], val[200000 + 5];
void solve(long long cur, long long depth) {
  if (depth % 2 == 0) {
    sum[cur] = INF;
    for (long long x : children[cur]) {
      sum[cur] = ((sum[cur]) < (sum[x]) ? (sum[cur]) : (sum[x]));
    }
    if (sum[cur] == INF) sum[cur] = sum[parent[cur]];
  }
  val[cur] = sum[cur] - sum[parent[cur]];
  for (long long x : children[cur]) solve(x, depth + 1);
}
int32_t main() {
  long long n;
  cin >> n;
  for (long long i = 2; i <= n; i++) {
    long long p;
    scanf("%I64d", &p);
    ;
    parent[i] = p;
    children[p].push_back(i);
  }
  for (long long i = 1; i <= n; i++) scanf("%I64d", &sum[i]);
  ;
  ;
  val[1] = sum[1];
  for (long long x : children[1]) {
    solve(x, 2);
  };
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    if (val[i] < 0) {
      cout << -1 << endl;
      return 0;
    }
    ans += val[i];
  }
  cout << ans << endl;
  return 0;
}
