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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const long long INF = 1000000000000000000LL;
const int MAX_N = 100005;
int n;
int p[MAX_N];
vector<int> g[MAX_N];
long long s[MAX_N];
long long result = 0;
bool haveResult = true;
void dfs(int u) {
  if (u == 0)
    result += s[u];
  else if (s[u] == -1) {
    if (!g[u].empty()) {
      long long curr = INF;
      for (int v : g[u]) curr = min(curr, s[v] - s[p[u]]);
      if (curr < 0)
        haveResult = false;
      else {
        result += curr;
        for (int v : g[u]) result += s[v] - curr - s[p[u]];
      }
    }
  }
  for (int v : g[u]) dfs(v);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i < n; i++) {
    cin >> p[i];
    p[i]--;
    g[p[i]].push_back(i);
  }
  for (int i = 0; i < n; i++) cin >> s[i];
  dfs(0);
  cout << (haveResult ? result : -1) << endl;
}
