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 infinity = 8e18;
const int mod = 998244353;
struct edge {
  int v;
  int connect;
} g[200004];
int h[200004];
int total;
int n, m;
long long s[200004];
void addedge(int u, int v) {
  g[total].v = v;
  g[total].connect = h[u];
  h[u] = total;
  ++total;
}
long long a[200004];
int check;
void dfsTraverse1(int u) {
  if (s[u] == -1) {
    long long minimum = infinity;
    for (int i = h[u]; i != -1; i = g[i].connect) {
      int v = g[i].v;
      dfsTraverse1(v);
      minimum = min(minimum, s[v]);
    }
    s[u] = minimum;
  } else {
    for (int i = h[u]; i != -1; i = g[i].connect) {
      int v = g[i].v;
      dfsTraverse1(v);
    }
  }
}
void dfsTraverse2(int u, int w) {
  if (s[u] == infinity) s[u] = w;
  a[u] = s[u] - w;
  if (a[u] < 0) {
    check = 0;
    return;
  }
  for (int i = h[u]; i != -1; i = g[i].connect) {
    int v = g[i].v;
    dfsTraverse2(v, s[u]);
  }
}
int main() {
  memset(h, -1, sizeof(h));
  total = 0;
  int n;
  cin >> n;
  int p;
  for (int i = 2; i <= n; ++i) {
    cin >> p;
    addedge(p, i);
  }
  check = 1;
  for (int i = 1; i <= n; ++i) {
    cin >> s[i];
  }
  dfsTraverse1(1);
  dfsTraverse2(1, 0);
  if (check) {
    long long minsum = 0;
    for (int i = 1; i <= n; ++i) {
      minsum += a[i];
    }
    cout << minsum;
  } else
    cout << "-1";
  return 0;
}
