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 long long MXN = 2e5 + 10;
long long n;
long long A[MXN], B[MXN];
vector<long long> adj[MXN];
void dfs(long long u, long long par, long long h) {
  if (par) {
    if (h % 2) {
      A[u] = -B[par] + B[u];
    } else {
      A[u] = 0;
      B[u] = B[par] + A[u];
    }
  }
  for (auto v : adj[u]) {
    if (v != par) dfs(v, u, h + 1);
  }
}
void ok(long long u, long long par, long long h) {
  for (auto v : adj[u]) {
    if (v != par) ok(v, u, h + 1);
  }
  long long mini = -1;
  for (auto v : adj[u]) {
    if (v != par) {
      mini = ((mini == -1 || A[v] < mini) ? A[v] : mini);
    }
  }
  if (mini > 0 && h % 2 == 0) {
    A[u] += mini;
    for (auto v : adj[u]) {
      if (v != par) {
        A[v] -= mini;
      }
    }
  }
}
void solve() {
  cin >> n;
  for (int i = 2; i <= n; i++) {
    long long p;
    cin >> p;
    adj[i].push_back(p), adj[p].push_back(i);
  }
  for (int i = 1; i <= n; i++) cin >> B[i];
  A[1] = B[1];
  dfs(1, 0, 1);
  ok(1, 0, 1);
  if (*min_element(A + 1, A + n + 1) < 0) {
    cout << -1 << '\n';
    return;
  }
  long long sum = 0;
  for (int i = 1; i <= n; i++) sum += A[i];
  cout << sum << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long q = 1;
  while (q--) solve();
  return 0;
}
