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;
void test() { cerr << "\n"; }
template <typename T, typename... Args>
void test(T x, Args... args) {
  cerr << x << " ";
  test(args...);
}
const int MAXN = (int)1e5 + 5;
const int MOD = (int)1e9 + 7;
void up(int &x, int y) {
  if (y == -1)
    return;
  else if (x == -1)
    x = y;
  else
    x = min(x, y);
}
int s[MAXN], a[MAXN], p[MAXN];
bool f[MAXN];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) scanf("%d", &p[i]);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &s[i]);
    f[i] = s[i] == -1;
  }
  for (int i = n; i > 1; i--)
    if (f[p[i]]) up(s[p[i]], s[i]);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    if (s[i] == -1) continue;
    int vp = a[p[i]];
    if (s[i] < vp) {
      ans = -1;
      break;
    }
    a[i] = s[i] - vp;
    ans += a[i];
    a[i] += a[p[i]];
  }
  printf("%lld\n", ans);
  return 0;
}
