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;
long long power(long long x, long long y, long long md) {
  long long res = 1;
  x %= md;
  while (y > 0) {
    if (y & 1) res = (res * x) % md;
    x = (x * x) % md;
    y = y >> 1;
  }
  return res % md;
}
vector<long long> v[100010];
vector<long long> second(100010);
vector<long long> a(100010, 0);
vector<bool> visit(100010, false);
void dfs(long long x, long long presum) {
  if (visit[x]) return;
  visit[x] = true;
  if (second[x] != -1) {
    second[x] -= presum;
  }
  for (long long i = 0; i < v[x].size(); ++i) {
    if (second[x] != -1) {
      dfs(v[x][i], second[x] + presum);
    } else {
      dfs(v[x][i], presum);
    }
  }
}
int main() {
  long long n;
  cin >> n;
  for (long long i = 0; i < n - 1; ++i) {
    long long x;
    cin >> x;
    v[x].push_back(i + 2);
  }
  vector<long long> t(n + 1, 0);
  for (long long i = 0; i < n; ++i) {
    cin >> second[i + 1];
    t[i + 1] = second[i + 1];
  }
  long long ans = second[1];
  dfs(1, 0);
  bool flag = false;
  for (long long i = 0; i < n; ++i) {
    long long mn = 1e15;
    if (second[i + 1] == -1) {
      for (long long j = 0; j < v[i + 1].size(); ++j) {
        mn = min(mn, second[v[i + 1][j]]);
      }
      for (long long j = 0; j < v[i + 1].size(); ++j) {
        ans += (second[v[i + 1][j]] - mn);
      }
      if (mn != 1e15) ans += mn;
    }
    if (t[i + 1] != -1 && second[i + 1] < 0) {
      flag = true;
      break;
    }
  }
  if (flag) ans = -1;
  cout << ans;
}
