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;
double pi = 2 * acos(0.0);
const long long int MOD = 1e9 + 7;
const long long int INF = 1e9 + 7;
const long long int MAX = 1e18;
const long long int N = 200005;
long long int n;
vector<long long int> adj[N];
vector<long long int> sum(N);
long long int ans = 0;
bool flag = true;
void dfs(long long int src, long long int par) {
  long long int mini = INT_MAX;
  for (auto x : adj[src]) {
    if (x == par) continue;
    mini = min(mini, sum[x]);
    dfs(x, src);
  }
  if (mini != INT_MAX && mini != -1 && sum[par] != -1 && sum[par] > mini) {
    flag = false;
    return;
  }
  if (sum[src] == -1) {
    if (mini == INT_MAX)
      sum[src] = sum[par];
    else
      sum[src] = mini;
  }
}
void dfs2(long long int src, long long int par) {
  ans += sum[src] - sum[par];
  for (auto x : adj[src]) {
    if (x == par) continue;
    dfs2(x, src);
  }
}
void WetPitch() {
  cin >> n;
  flag = true;
  for (long long int i = 2; i <= n; i++) {
    long long int x;
    cin >> x;
    adj[i].push_back(x);
    adj[x].push_back(i);
  }
  for (long long int i = 1; i <= n; i++) {
    cin >> sum[i];
  }
  dfs(1, 0);
  if (flag == false) {
    cout << -1 << '\n';
    return;
  }
  dfs2(1, 0);
  cout << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int tst;
  tst = 1;
  while (tst--) {
    WetPitch();
  }
  return 0;
}
