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> st[100000 + 5];
long long vis[100000 + 5];
long long a[100000 + 5];
int last[100000 + 5];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    int x;
    scanf("%d", &x);
    st[x].push_back(i);
    last[i] = x;
  }
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    vis[i] = x;
  }
  queue<int> stt;
  stt.push(1);
  int flag = 1;
  a[1] = vis[1];
  while (!stt.empty()) {
    int temp = stt.front();
    stt.pop();
    long long min0 = 1000000000000000005;
    int l = st[temp].size();
    if (l == 0 && vis[temp] == -1) continue;
    for (int i = 0; i < l; i++) {
      int x = st[temp][i];
      min0 = min(min0, vis[x]);
      stt.push(x);
    }
    if (temp == 1) continue;
    if (vis[temp] == -1) {
      if (vis[last[temp]] > min0)
        flag = 0;
      else {
        vis[temp] = min0;
        a[temp] = min0 - vis[last[temp]];
      }
    } else {
      a[temp] = vis[temp] - vis[last[temp]];
    }
  }
  if (flag == 0) {
    printf("-1\n");
    return 0;
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) ans += a[i];
  printf("%lld\n", ans);
}
