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 maxn = 1e5 + 100;
int num[maxn];
vector<int> mp[maxn];
int flag;
long long ans = 0;
int dfs1(int id, int fa) {
  if (mp[id].size() == 0) {
    if (num[id] == -1) {
      num[id] = num[fa];
    }
    return num[id];
  }
  int v = 0x3f3f3f3f;
  for (int i = 0; i < mp[id].size(); i++) {
    v = min(v, dfs1(mp[id][i], id));
  }
  if (num[id] == -1) num[id] = v;
  return num[id];
}
void dfs2(int id, int fa) {
  if (num[id] < num[fa]) {
    flag = -1;
    return;
  }
  ans += num[id] - num[fa];
  if (mp[id].size() == 0) {
    return;
  }
  for (int i = 0; i < mp[id].size(); i++) {
    dfs2(mp[id][i], id);
  }
  return;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    int fa;
    scanf("%d", &fa);
    mp[fa].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &num[i]);
  }
  dfs1(1, 0);
  dfs2(1, 0);
  if (flag == -1)
    printf("%d\n", flag);
  else
    printf("%lld\n", ans);
  return 0;
}
