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 MX = 100005;
long long n, p[MX], s[MX], visited[MX], ans, chk;
vector<int> v[MX];
void dfs(int x, long long val) {
  visited[x] = 1;
  long long u = 99999999999999999;
  if (s[x] == -1) {
    for (int i = 0; i < v[x].size(); i++) {
      int next = v[x][i];
      if (!visited[next] && s[next] != -1) {
        u = min(u, s[next]);
        if (val > s[next]) chk = 1;
      }
    }
    if (u == 99999999999999999)
      s[x] = val;
    else
      s[x] = u;
    ans += s[x] - val;
    val = s[x];
  } else {
    if (s[x] < val) chk = 1;
    ans += s[x] - val;
    val = s[x];
  }
  for (int i = 0; i < v[x].size(); i++) {
    int next = v[x][i];
    if (!visited[next]) {
      dfs(next, val);
    }
  }
}
int main() {
  cin >> n;
  for (int i = 2; i <= n; i++) {
    int x;
    cin >> x;
    v[x].push_back(i);
    v[i].push_back(x);
  }
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
  }
  dfs(1, 0);
  if (!chk)
    cout << ans;
  else
    cout << -1;
}
