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 maxn = 3e5 + 5;
const int mod = 1e9 + 7;
long long qpow(long long a, long long b) {
  long long ans = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
  }
  return ans;
}
long long gcd(long long a, long long b) { return b > 0 ? gcd(b, a % b) : a; }
int n, m, T;
long long a[maxn];
vector<int> g[maxn];
int pre[maxn];
long long mi[maxn];
int ans[maxn];
bool dfs(int u, int fa) {
  if (u != 1 && a[u] == -1) {
    if (g[u].size() == 0)
      a[u] = a[fa];
    else
      a[u] = mi[u];
  }
  if (a[fa] > a[u]) return 0;
  ans[u] = a[u] - a[fa];
  for (auto v : g[u]) {
    if (dfs(v, u) == 0) return 0;
  }
  return 1;
}
int main() {
  cin >> n;
  for (int i = 2; i <= n; i++)
    cin >> pre[i], g[pre[i]].push_back(i), mi[i] = mod;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] != -1) mi[pre[i]] = min(mi[pre[i]], a[i]);
  }
  if (dfs(1, 0) == 0) {
    puts("-1");
    return 0;
  }
  ans[1] = a[1];
  long long sum = 0;
  for (int i = 1; i <= n; i++) {
    sum += ans[i];
  }
  cout << sum;
  return 0;
}
