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> g[100005];
int vis[100005];
int main() {
  int n;
  cin >> n;
  int p[n + 2];
  int a[n + 2];
  int s[n + 2];
  for (int i = 2; i <= n; i++) cin >> p[i];
  p[1] = p[2];
  cin >> s[1];
  for (int i = 2; i <= n; i++) {
    cin >> s[i];
    g[p[i]].push_back(i);
    g[i].push_back(p[i]);
  }
  s[0] = 0;
  queue<pair<int, int> > q;
  q.push({1, 0});
  vis[1] = 1;
  int f = 0;
  long long ans = 0;
  while (!q.empty()) {
    int x = q.front().first;
    int y = q.front().second;
    vis[x] = 1;
    int mx = INT_MAX;
    int c = 0;
    for (auto y : g[x]) {
      if (vis[y]) continue;
      mx = min(mx, s[y]);
      q.push({y, x});
      c = 1;
    }
    if (!c)
      s[x] = max(max(0, s[y]), s[x]);
    else if (s[x] == -1)
      s[x] = mx;
    else if (s[x] > mx && mx != -1)
      f = 1;
    ans += (s[x] - s[y]);
    if (s[x] < s[y]) f = 1;
    q.pop();
  }
  if (f)
    cout << -1;
  else
    cout << ans;
  printf("\n");
}
