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 = 100010;
long long ans;
vector<int> g[N];
bool flag;
int n, p, a[N], s[N];
void dfs(int v, int p) {
  if (s[v] == -1) {
    s[v] = (int)(g[v]).size() ? 0x3f3f3f3f : s[p];
    for (auto& u : g[v]) s[v] = min(s[v], s[u]);
  }
  a[v] = s[v] - s[p];
  for (auto& u : g[v]) dfs(u, v);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (int i = (2); i < (n + 1); i++) {
    cin >> p;
    g[p].emplace_back(i);
  }
  for (int i = (1); i < (n + 1); i++) cin >> s[i];
  dfs(1, 0);
  for (int i = (1); i < (n + 1); i++) {
    if (a[i] < 0) flag = true;
    ans += a[i];
  }
  cout << (flag ? -1 : ans) << '\n';
}
