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 N = 200 * 1000 + 10;
const long long MOD = 1000000007;
const long long INF = 10000000000;
long long binpow(long long a, long long n) {
  if (n == 0) return 1;
  if (n % 2) {
    return (binpow(a, n - 1) * a) % MOD;
  } else {
    long long b = binpow(a, n / 2);
    return (b * b) % MOD;
  }
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long n, p[N], s[N], a[N], ans;
vector<long long> v[N];
void bfs(long long k, long long sum) {
  if (v[k].size() == 0) {
    if (s[k] == -1) {
      a[k] = 0;
    } else {
      a[k] = s[k] - sum;
    }
    return;
  }
  if (s[k] == -1) {
    long long m = INF;
    for (int i = 0; i < (int)v[k].size(); ++i) {
      m = min(m, s[v[k][i]]);
    }
    a[k] = m - sum;
    sum += a[k];
  } else {
    a[k] = s[k] - sum;
    sum += a[k];
  }
  for (int i = 0; i < (int)v[k].size(); ++i) {
    bfs(v[k][i], sum);
  }
}
int main(void) {
  cin >> n;
  for (long long i = 2; i <= n; ++i) {
    cin >> p[i];
    v[p[i]].push_back(i);
  }
  for (long long i = 1; i <= n; ++i) {
    cin >> s[i];
  }
  bfs(1, 0);
  for (int i = 1; i <= n; ++i) {
    if (a[i] < 0) {
      cout << -1;
      return 0;
    }
    ans += a[i];
  }
  cout << ans;
  return 0;
}
