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;
int n;
vector<int> lst[100005];
long long m[100005];
long long cur[100005], sum[100005];
long long val[100005], res;
const long long INF = 1e9 + 7;
long long dfs(int v) {
  long long ret = INF;
  ret = min(ret, m[v]);
  for (int i = 0; i < lst[v].size(); i++) {
    int nxt = lst[v][i];
    ret = min(ret, dfs(nxt));
  }
  val[v] = ret;
  return ret;
}
void pro(int v) {
  long long ret = INF;
  if (m[v] == INF) {
    if (val[v] != INF) {
      if (val[v] - cur[v] >= 0) {
        res += val[v] - cur[v];
        cur[v] = val[v];
      } else {
        printf("-1");
        exit(0);
      }
    }
  } else {
    if (m[v] - cur[v] < 0) {
      printf("-1");
      exit(0);
    }
    res += m[v] - cur[v];
    cur[v] = m[v];
  }
  for (int i = 0; i < lst[v].size(); i++) {
    int nxt = lst[v][i];
    cur[nxt] = cur[v];
    pro(nxt);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    int x;
    scanf("%d", &x);
    lst[x].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &m[i]);
    if (m[i] == -1) m[i] = INF;
  }
  dfs(1);
  pro(1);
  printf("%lld", res);
  return 0;
}
