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 maxk = 18;
const int maxn = 2e5 + 5;
int N;
long long in[maxn];
long long ans[maxn];
vector<int> graph[maxn];
void DFS(int u, int p, int lvl, long long sum) {
  if (lvl) {
    if (in[u] != -1) {
      cout << -1 << endl;
      exit(0);
    }
    long long minx = 1LL << 51;
    for (int v : graph[u])
      if (v != p) minx = min(minx, in[v] - sum);
    if (minx < 0) {
      cout << -1 << endl;
      exit(0);
    }
    if (minx == (1LL << 51)) minx = 0;
    ans[u] = minx;
  }
  if (!lvl) {
    if (sum > in[u]) {
      cout << -1 << endl;
      exit(0);
    }
    ans[u] = in[u] - sum;
  }
  for (int v : graph[u])
    if (v != p) DFS(v, u, lvl ^ 1, sum + ans[u]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> N;
  for (int i = 2; i <= N; i++) {
    int x;
    cin >> x;
    graph[i].push_back(x);
    graph[x].push_back(i);
  }
  for (int i = 1; i <= N; i++) cin >> in[i];
  DFS(1, 1, 0, 0);
  long long out = 0;
  for (int i = 1; i <= N; i++) out += ans[i];
  cout << out << endl;
  return 0;
}
