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;
vector<int> g[200309];
int n, par[200309];
long long s[200309], a[200309];
bool ok = true;
void dfs(int u) {
  if (s[u] == -1) {
    if (g[u].empty())
      a[u] = 0;
    else {
      a[u] = 0x3f3f3f3f3f3f3f3f;
      for (int v : g[u]) {
        a[u] = min(a[u], s[v] - s[par[u]]);
      }
      s[u] = a[u] + s[par[u]];
    }
  } else {
    a[u] = s[u] - s[par[u]];
  }
  if (a[u] < 0) ok = false;
  for (int v : g[u]) dfs(v);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; (i) <= int(n - 1); (i)++) {
    int p;
    scanf("%d", &p);
    g[p].push_back(i + 1);
    par[i + 1] = p;
  }
  par[1] = 0;
  s[0] = 0;
  for (int i = 1; (i) <= int(n); (i)++) {
    scanf("%lld", &s[i]);
  }
  dfs(1);
  long long ans = 0;
  for (int u = 1; (u) <= int(n); (u)++) ans += a[u];
  if (!ok) ans = -1;
  printf("%lld\n", ans);
  return 0;
}
