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 = 200005;
stack<char> stk;
stack<long long> stk1;
map<char, int> mp;
set<int> seet;
vector<pair<string, int> > Pi;
vector<string> v, v1;
vector<int> graph[maxn];
int depth[maxn];
long long arr[maxn];
long long Ans, pos = 11;
int POW(int n) {
  int s = 1;
  for (int i = 0; i < n; i++) s *= 2;
  return s;
}
void Depth(int u, int par, int dep) {
  depth[u] = dep;
  for (auto v : graph[u])
    if (par != v) {
      Depth(v, u, dep + 1);
    }
}
void DFS(long long u, long long par, long long sum) {
  long long mx = 1e18;
  if (par == -1) {
    Ans += sum;
    for (auto v : graph[u]) {
      if (par != v) DFS(v, u, sum);
    }
  } else {
    if (arr[u] == -1) {
      for (auto v : graph[u]) {
        if (par != v) {
          if (sum > arr[v]) {
            pos = 0;
          } else
            mx = min(mx, arr[v] - sum);
        }
      }
    }
    if (arr[u] == -1) {
      if (mx != 1e18) Ans += mx;
      for (auto v : graph[u]) {
        if (par != v) DFS(v, u, sum + mx);
      }
    } else {
      if (sum > arr[u])
        pos = 0;
      else
        Ans += (arr[u] - sum);
      for (auto v : graph[u]) {
        if (par != v) DFS(v, u, arr[u]);
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  long long n;
  cin >> n;
  for (long long i = 2; i <= n; i++) {
    long long u;
    cin >> u;
    graph[u].push_back(i);
    graph[i].push_back(u);
  }
  for (long long i = 1; i <= n; i++) cin >> arr[i];
  Depth(1, -1, 1);
  for (int i = 1; i <= n; i++) {
    if (depth[i] % 2 == 0 and arr[i] == -1) continue;
    if (depth[i] % 2 == 1 and arr[i] != -1) continue;
    pos = 0;
  }
  DFS(1, -1, arr[1]);
  if (pos == 0)
    cout << -1 << endl;
  else
    cout << Ans << endl;
  return 0;
}
