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 = 2e5 + 50;
const int M = 1e6;
const long long oo = 1e18;
const int mod = 1e9 + 7;
vector<int> g[N];
long long s[N], ans[N];
int a[N];
int d[N];
int dfs(int u, int p) {
  int b = 2e9;
  d[u] = d[p] + 1;
  if (d[u] % 2) b = s[u];
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    int tmp = dfs(v, u);
    b = min(b, tmp);
  }
  a[u] = b;
  return b;
}
void dfs2(int u, int cur) {
  if (d[u] % 2 && cur > s[u] || d[u] % 2 == 0 && cur > a[u]) {
    puts("-1");
    exit(0);
  }
  int dif = 0;
  if (d[u] % 2 == 0 && g[u].size()) {
    dif = a[u] - cur;
    ans[u] = dif;
  } else if (d[u] % 2)
    ans[u] = s[u] - cur;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    dfs2(v, cur + ans[u]);
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 2; i <= n; i++) {
    int p;
    cin >> p;
    g[p].push_back(i);
  }
  for (int i = 1; i <= n; i++) cin >> s[i];
  dfs(1, 0);
  dfs2(1, 0);
  long long res = 0;
  for (int i = 1; i <= n; i++) res += ans[i];
  cout << res << endl;
}
