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 int INF = 1e18;
const long long int N = 1e5 + 5;
long long int mod = 998244353;
long long int dfs(long long int node, long long int par,
                  vector<vector<long long int>> &g, bool &flag,
                  vector<long long int> &sum, vector<long long int> &nsum) {
  if ((long long int)g[node].size() == 1 and node != 1) {
    if (sum[node] == -1) {
      return nsum[node] = INF;
    } else {
      return nsum[node] = sum[node];
    }
  }
  long long int mina = INF;
  for (auto &nbr : g[node]) {
    if (nbr == par) continue;
    mina = min(dfs(nbr, node, g, flag, sum, nsum), mina);
  }
  if (mina < sum[node]) {
    flag = true;
  }
  if (sum[node] != -1) {
    mina = min(mina, sum[node]);
  }
  return nsum[node] = mina;
}
void dfs1(long long int node, long long int par, long long int prev,
          vector<long long int> &nsum, vector<vector<long long int>> &g,
          long long int &ans) {
  if (nsum[node] != INF) ans += nsum[node] - prev;
  for (auto &nbr : g[node]) {
    if (nbr == par) continue;
    dfs1(nbr, node, nsum[node], nsum, g, ans);
  }
}
void null() {
  long long int n;
  cin >> n;
  vector<vector<long long int>> g(n + 1);
  for (long long int i = 2; i < n + 1; i++) {
    long long int x;
    cin >> x;
    g[i].push_back(x);
    g[x].push_back(i);
  }
  vector<long long int> sum(n + 1, 0), nsum(n + 1, 0);
  for (long long int i = 1; i < n + 1; i++) {
    cin >> sum[i];
  }
  long long int ans = 0;
  bool flag = false;
  dfs(1, 0, g, flag, sum, nsum);
  if (flag) {
    cout << -1 << '\n';
    return;
  }
  dfs1(1, 0, 0, nsum, g, ans);
  cout << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(0);
  cout << setprecision(4);
  cout << fixed;
  long long int t = 1;
  clock_t start, end;
  start = clock();
  while (t--) {
    null();
  }
  end = clock();
  double time_taken = double(end - start) / double(CLOCKS_PER_SEC);
}
