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 MAXN = 2e5 + 7;
long long used[MAXN];
vector<vector<long long>> g;
vector<long long> s(MAXN);
vector<long long> a(MAXN, -1);
vector<long long> p(MAXN);
long long ans = 0;
void dfs(long long u) {
  if (s[u] == -1) {
    s[u] = 1e18;
    for (auto v : g[u]) {
      if (!used[v]) {
        s[u] = min(s[u], s[v]);
      }
    }
    if (s[u] == 1e18) {
      s[u] = s[p[u]];
    }
  }
  if (u != 0) {
    a[u] = s[u] - s[p[u]];
    if (a[u] < 0) {
      ans = -1e18;
    } else {
      ans += a[u];
    }
  } else {
    ans += s[0];
  }
  used[u] = true;
  for (auto v : g[u]) {
    if (!used[v]) {
      dfs(v);
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n;
  cin >> n;
  g.resize(n);
  for (long long i = 0; i < n - 1; ++i) {
    long long v;
    cin >> v;
    --v;
    p[i + 1] = v;
    g[i + 1].push_back(v);
    g[v].push_back(i + 1);
  }
  for (long long i = 0; i < n; ++i) {
    cin >> s[i];
  }
  a[0] = s[0];
  dfs(0);
  if (ans < 0) {
    cout << -1 << endl;
  } else {
    cout << ans << endl;
  }
  return 0;
}
