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 long long MOD = 1000000007;
const long long MX = 100005;
long long fpow(long long x, long long y) {
  if (y == 0) return 1;
  if (y == 1) return x;
  long long p = fpow(x, y / 2);
  if (y % 2 != 0) p *= p;
  return p;
}
long long GCD(long long x, long long y) {
  if (y == 0) return x;
  return GCD(y, x % y);
}
bool vis[MX], b;
long long a[MX];
vector<long long> v[MX];
long long sum = 0;
void solve(int ver, long long cur, int dep, int par) {
  long long Do = cur;
  if (vis[ver]) return;
  vis[ver] = 1;
  if ((dep % 2) == 1) {
    sum += a[ver] - cur;
    Do = a[ver];
  } else {
    long long x = 0, y = -1;
    for (int i = 0; i < v[ver].size(); i++) {
      if (v[ver][i] == par) continue;
      if (a[v[ver][i]] < cur) b = 1;
      if (y == -1) {
        y = a[v[ver][i]];
      }
      y = min(y, a[v[ver][i]]);
    }
    if (y != -1) y -= cur;
    sum += max(y, x);
    Do = cur + max(x, y);
  }
  for (int i = 0; i < v[ver].size(); i++) {
    solve(v[ver][i], Do, dep + 1, ver);
  }
  return;
}
int main() {
  int n;
  cin >> n;
  for (int i = 2; i <= n; i++) {
    int x;
    cin >> x;
    v[i].push_back(x);
    v[x].push_back(i);
  }
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  solve(1, 0, 1, -12);
  if (b) return cout << -1, 0;
  cout << sum;
  return 0;
}
