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;
long long N;
vector<long long> adj[100005];
long long s[100005], a[100005];
bool fu;
void dfs(long long u, long long sum, long long h) {
  if (fu) return;
  long long t = adj[u].size();
  long long me = 0x3F3F3F3F;
  if (h % 2 == 0) {
    for (int i = int(0); i < int(t); i++) {
      long long v = adj[u][i];
      me = min(me, s[v]);
    }
  }
  for (int i = int(0); i < int(t); i++) {
    long long v = adj[u][i];
    if (h & 1) {
      if (u == 1)
        dfs(v, sum + a[u], h + 1);
      else {
        a[u] = s[u] - sum;
        dfs(v, sum + a[u], h + 1);
      }
    } else {
      if (me < sum) {
        fu = 1;
        break;
      } else {
        long long need = me - sum;
        a[u] = need;
        dfs(v, sum + a[u], h + 1);
      }
    }
  }
  if (t == 0 && (h & 1)) {
    a[u] = s[u] - sum;
  }
}
int main() {
  cin >> N;
  for (int i = int(2); i < int(N + 1); i++) {
    int x;
    scanf("%d", &x);
    adj[x].push_back(i);
  }
  for (int i = int(1); i < int(N + 1); i++) scanf("%d", &s[i]);
  a[1] = s[1];
  fu = false;
  dfs(1, 0, 1);
  if (fu)
    cout << "-1\n";
  else {
    long long ans = 0;
    for (int i = 1; i <= N; i++) {
      ans += a[i];
    }
    cout << ans << endl;
  }
  return 0;
}
