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 h[1000000], a, i, n, j, s[1000004], v[1000000], p[1000000];
long long sum = 0, d, b, mini, sum1, e;
int main() {
  vector<long long> vec[100004];
  h[1] = 1;
  scanf("%lld", &n);
  for (i = 2; i <= n; i++) {
    scanf("%lld", &a);
    h[i] = h[a] + 1;
    vec[a].push_back(i);
    p[i] = a;
  }
  for (i = 1; i <= n; i++) {
    scanf("%lld", &s[i]);
  }
  for (i = 1; i <= n; i++) {
    if (i == 1) {
      v[i] = s[i];
    } else {
      if (s[i] != -1) {
        sum1 = 0;
        d = 0;
        a = s[i];
        b = p[i];
        while (1) {
          if (s[b] == -1) {
            sum1 = sum1 + v[b];
            b = p[b];
            continue;
          } else {
            if (a >= s[b]) {
              v[i] = a - (s[b] + sum1);
              break;
            } else {
              d = 1;
              break;
            }
          }
        }
        if (d == 1) {
          break;
        }
      } else if (s[i] == -1) {
        e = 0;
        d = 0;
        a = s[p[i]];
        mini = 100000000000;
        for (j = 0; j < vec[i].size(); j++) {
          e++;
          if (s[vec[i][j]] >= a) {
            b = s[vec[i][j]] - a;
            if (b < mini) {
              mini = b;
            }
          } else {
            d = 1;
            break;
          }
        }
        if (e > 0) {
          v[i] = mini;
        } else {
          v[i] = 0;
        }
      }
    }
  }
  if (d == 1) {
    printf("-1\n");
  } else {
    for (i = 1; i <= n; i++) {
      sum = sum + v[i];
    }
    printf("%lld\n", sum);
  }
}
