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> q[100010];
const long long inf = 10000000001;
long long s[100010];
struct node {
  int dep, id;
  bool operator<(const node& aa) const { return aa.dep > dep; }
} a[100010];
void caul(int st, int deep) {
  a[st].dep = deep;
  a[st].id = st;
  int si = q[st].size();
  for (int i = 0; i < si; i++) {
    caul(q[st][i], deep + 1);
  }
}
long long ans[100010];
int fa[100010];
int main() {
  int n, f;
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    scanf("%d", &f);
    fa[i] = f;
    q[f].push_back(i);
  }
  for (int i = 1; i <= n; i++) scanf("%lld", &s[i]);
  caul(1, 1);
  sort(a + 1, a + n + 1);
  ans[1] = s[1];
  int flag = 1;
  for (int i = 1; i <= n; i++) {
    if ((a[i].dep) & 1) {
      int si = q[a[i].id].size();
      for (int j = 0; j < si; j++) {
        s[q[a[i].id][j]] = s[a[i].id];
      }
    } else {
      int si = q[a[i].id].size();
      if (si == 0)
        ans[a[i].id] = 0;
      else {
        long long minn = inf;
        for (int j = 0; j < si; j++) {
          if (s[q[a[i].id][j]] < s[a[i].id]) {
            flag = 0;
            break;
          }
          minn = min(minn, s[q[a[i].id][j]]);
        }
        ans[a[i].id] = minn - s[a[i].id];
        s[a[i].id] += ans[a[i].id];
        for (int j = 0; j < si; j++) {
          ans[q[a[i].id][j]] = s[q[a[i].id][j]] - s[a[i].id];
        }
      }
    }
    if (flag == 0) break;
  }
  if (flag == 0) {
    printf("-1\n");
  } else {
    long long ans1 = 0;
    for (int i = 1; i <= n; i++) {
      ans1 += ans[i];
    }
    printf("%lld\n", ans1);
  }
  return 0;
}
