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 inf = LONG_MAX;
const long long arr = 1000000;
bool comp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first == b.first)
    return a.second < b.second;
  else
    return a.first < b.first;
}
vector<vector<long long> > g;
bool first = false;
vector<bool> used(arr, 0);
vector<long long> a(arr, 0);
vector<long long> vec(arr, 0);
void dfs(long long v, long long prev, long long sum) {
  used[v] = true;
  if (a[v] != -1)
    vec[v] = a[v] - sum;
  else
    vec[v] = 0;
  if (vec[v] < 0) first = true;
  if (a[v] != -1) sum = a[v];
  for (int i = 0; i < g[v].size(); i++) {
    long long to = g[v][i];
    if (to == prev) continue;
    if (!used[to]) {
      dfs(to, v, sum);
    }
  }
}
void dfs2(long long v, long long prev) {
  used[v] = true;
  long long mn = LLONG_MAX;
  for (int i = 0; i < g[v].size(); i++) {
    long long to = g[v][i];
    if (to == prev) continue;
    mn = min(mn, vec[to]);
  }
  if (mn == LLONG_MAX) mn = 0;
  vec[v] += mn;
  for (int i = 0; i < g[v].size(); i++) {
    long long to = g[v][i];
    if (to == prev) continue;
    vec[to] -= mn;
  }
  for (int i = 0; i < g[v].size(); i++) {
    long long to = g[v][i];
    if (to == prev) continue;
    if (!used[to]) {
      dfs2(to, v);
    }
  }
}
int main(void) {
  long long n;
  cin >> n;
  g.resize(n + 1);
  for (long long i = 2; i <= n; i++) {
    long long x;
    cin >> x;
    g[x].push_back(i);
    g[i].push_back(x);
  }
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  dfs(1, -1, 0);
  if (first) return cout << -1 << '\n', 0;
  for (int i = 0; i < arr; i++) used[i] = false;
  dfs2(1, -1);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    ans += vec[i];
  }
  cout << ans << '\n';
  return 0;
}
