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 = 1e18;
const int maxn = 1e5 + 10;
vector<int> adj[maxn];
long long sum[maxn], ans, flag;
void dfs(int u, long long pre, bool opt) {
  if (opt) {
    if (sum[u] < 0 || sum[u] - pre < 0) {
      flag = 0;
      return;
    }
    ans += sum[u] - pre;
    for (int i = 0; i < adj[u].size(); i++) dfs(adj[u][i], sum[u], !opt);
    return;
  }
  long long cur = inf;
  for (int i = 0; i < adj[u].size(); i++) cur = min(cur, sum[adj[u][i]] - pre);
  if (cur < 0) {
    flag = 0;
    return;
  }
  if (cur == inf) cur = 0;
  ans += cur;
  for (int i = 0; i < adj[u].size(); i++) dfs(adj[u][i], pre + cur, !opt);
}
int main() {
  int n;
  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("%lld", &sum[i]);
  flag = 1;
  dfs(1, 0, 1);
  if (!flag)
    printf("-1");
  else
    printf("%lld", ans);
  return 0;
}
