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;
const int N = 100000, INF = 0x3f3f3f3f;
int eo[N], *ej[N], ss[N];
void append(int i, int j) {
  int o = eo[i]++;
  if (o >= 2 && (o & o - 1) == 0)
    ej[i] = (int *)realloc(ej[i], (o << 1) * sizeof *ej[i]);
  ej[i][o] = j;
}
long long dfs(int p, int i, int h) {
  long long ans = 0;
  if (h % 2 == 0) {
    if (eo[i] == 1) return 0;
    int s = INF;
    for (int o = eo[i]; o--;) {
      int j = ej[i][o];
      if (j != p)
        if (ss[j] >= ss[p])
          s = min(s, ss[j] - ss[p]);
        else
          return -1;
    }
    ans += s;
    for (int o = eo[i]; o--;) {
      int j = ej[i][o];
      if (j != p) ans += ss[j] - ss[p] - s;
    }
    for (int o = eo[i]; o--;) {
      int j = ej[i][o];
      if (j != p) {
        long long ans_ = dfs(i, j, h + 1);
        if (ans_ != -1)
          ans += ans_;
        else
          return -1;
      }
    }
  } else {
    for (int o = eo[i]; o--;) {
      int j = ej[i][o];
      if (j != p) {
        long long ans_ = dfs(i, j, h + 1);
        if (ans_ != -1)
          ans += ans_;
        else
          return -1;
      }
    }
  }
  return ans;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) ej[i] = (int *)malloc(2 * sizeof *ej[i]);
  for (int j = 1; j < n; j++) {
    int i;
    cin >> i, i--;
    append(i, j), append(j, i);
  }
  for (int i = 0; i < n; i++) cin >> ss[i];
  long long ans = dfs(-1, 0, 1);
  if (ans == -1) {
    cout << "-1\n";
    return 0;
  }
  ans += ss[0];
  cout << ans << '\n';
  for (int i = 0; i < n; i++) free(ej[i]);
  return 0;
}
