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;
vector<long long> chi[100001];
long long val[100001];
bool f = 1;
long long ans;
void dfs(long long, long long, bool);
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  long long x;
  for (long long i = 2; i <= n; i++) {
    cin >> x;
    chi[x].push_back(i);
  }
  for (long long i = 1; i <= n; i++) {
    cin >> x;
    val[i] = x;
    if (val[i] == -1) val[i] = LLONG_MAX;
  }
  ans = val[1];
  dfs(1, 0, 1);
  cout << (f ? ans : -1);
}
void dfs(long long v, long long m, bool h) {
  if (val[v] < m) {
    f = 0;
    return;
  }
  if (h) {
    for (auto x : chi[v]) {
      dfs(x, val[v], h ^ 1);
    }
  } else {
    vector<long long> hehe;
    long long ma = LLONG_MAX;
    for (auto x : chi[v]) {
      hehe.push_back(val[x]);
      ma = min(ma, val[x]);
      dfs(x, m, h ^ 1);
    }
    if (ma != LLONG_MAX) ans += ma - m;
    for (auto x : hehe) {
      ans += x - ma;
    }
  }
}
