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 = 1e5 + 10, inf = 0x3f3f3f3f;
void cmin(int& x, int y) {
  if (x > y) {
    x = y;
  }
}
int n, a[N];
long long answer;
vector<int> adj[N];
void dfs(int u, int p, int d, int s) {
  if (d & 1) {
    answer += a[u] - s;
    s = a[u];
  } else {
    int minv = inf;
    for (auto v : adj[u]) {
      if (v != p) {
        cmin(minv, a[v]);
      }
    }
    if (minv < s) {
      puts("-1");
      exit(0);
    } else if (minv != inf) {
      answer += minv - s;
      s = minv;
    }
  }
  for (auto v : adj[u]) {
    if (v != p) {
      dfs(v, u, d + 1, s);
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; ++i) {
    int x;
    scanf("%d", &x);
    adj[x].push_back(i);
  }
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  dfs(1, 0, 1, 0);
  printf("%lld\n", answer);
  return 0;
}
