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;
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-ffloat-store")
#pragma GCC optimize("-fno-defer-pop")
long long int power(long long int a, long long int b, long long int m) {
  if (b == 0) return 1;
  if (b == 1) return a % m;
  long long int t = power(a, b / 2, m) % m;
  t = (t * t) % m;
  if (b & 1) t = ((t % m) * (a % m)) % m;
  return t;
}
long long int modInverse(long long int a, long long int m) {
  return power(a, m - 2, m);
}
long long int ar[300005], n, ans[300005], fl[300005];
vector<long long int> vc[300005];
void dfs(long long int i, long long int p_val, long long int dpt) {
  long long int nx;
  fl[i] = 1;
  if (dpt % 2 == 0) {
    long long int mn = 1e18;
    for (auto it : vc[i]) {
      if (fl[it]) continue;
      mn = min(mn, ar[it]);
    }
    if (mn == 1e18)
      ans[i] = 0;
    else {
      ans[i] = mn - p_val;
      nx = mn;
    }
  } else {
    ans[i] = ar[i] - p_val;
    nx = ar[i];
  }
  for (auto it : vc[i]) {
    if (fl[it]) continue;
    dfs(it, nx, dpt + 1);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int i, j, k, l;
  cin >> n;
  for (i = 2; i <= n; i++) {
    cin >> k;
    vc[k].push_back(i);
    vc[i].push_back(k);
  }
  for (i = 1; i <= n; i++) {
    cin >> ar[i];
  }
  dfs(1, 0, 1);
  long long int flg = 1, s = 0;
  for (i = 1; i <= n; i++) {
    if (ans[i] < 0) {
      flg = 0;
      break;
    }
    s += ans[i];
  }
  if (flg)
    cout << s << "\n";
  else
    cout << "-1"
         << "\n";
  return 0;
}
