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 INF = LLONG_MAX;
const long long SIZE = 4e5 + 5;
const long long mod = 1e9 + 7;
vector<long long> adj[SIZE];
long long par[SIZE], a[SIZE], ss[SIZE];
void dfs1(long long u, long long d) {
  long long i, v, x, y;
  if (d % 2 == 1) {
    x = par[u];
    y = par[x];
    a[x] = min(a[x], ss[u] - ss[y]);
  } else if (adj[u].size() == 0)
    a[u] = 0;
  for (i = 0; i < adj[u].size(); i++) {
    v = adj[u][i];
    dfs1(v, d + 1);
  }
}
void dfs2(long long u, long long d) {
  long long i, v, x, y;
  if (d % 2 == 1)
    a[u] = ss[u] - ss[par[u]];
  else
    ss[u] = a[u] + ss[par[u]];
  for (i = 0; i < adj[u].size(); i++) {
    v = adj[u][i];
    dfs2(v, d + 1);
  }
}
int main() {
  long long t = 0, n, x, y, z, i, j, k, g, p, q, ans = 0, sum = 0, c = 0;
  string s, s1, s2;
  scanf("%lld", &n);
  for (i = 2; i <= n; i++) {
    scanf("%lld", &x);
    par[i] = x;
    adj[x].push_back(i);
  }
  for (i = 1; i <= n; i++) scanf("%lld", &ss[i]);
  for (i = 1; i <= n; i++) a[i] = INF;
  dfs1(1, 1);
  dfs2(1, 1);
  for (i = 1; i <= n; i++) {
    if (a[i] < 0) {
      printf("-1\n");
      return 0;
    }
    sum += a[i];
  }
  printf("%lld\n", sum);
  return 0;
}
