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 = 9223372036854775807LL;
const int MAX_N = 1e+5;
vector<int> adj[MAX_N];
bool visited[MAX_N];
long long a[MAX_N], h[MAX_N], s[MAX_N], p[MAX_N];
bool ok = true;
void dfs(int u, int anc) {
  visited[u] = true;
  p[u] = anc;
  if (anc == -1) {
    h[u] = 1LL;
    a[u] = s[u];
  } else
    h[u] = h[anc] + 1LL;
  int l = adj[u].size();
  if (h[u] % 2 == 0) s[u] = s[p[u]];
  long long mini = INF;
  for (int i = 0; i < l; i++) {
    int v = adj[u][i];
    if (not visited[v]) {
      dfs(v, u);
      if (h[u] % 2 == 0) mini = min(mini, s[v]);
    }
  }
  if (mini != INF) {
    if (s[u] > mini)
      ok = false;
    else
      s[u] = mini;
  }
}
void dfs(int u) {
  visited[u] = true;
  int l = adj[u].size();
  if (u != 0) a[u] = s[u] - s[p[u]];
  for (int i = 0; i < l; i++) {
    int v = adj[u][i];
    if (not visited[v]) dfs(v);
  }
}
void solve() {
  int n;
  cin >> n;
  for (int u = 1; u < n; u++) {
    int v;
    cin >> v;
    v--;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  for (int i = 0; i < n; i++) cin >> s[i];
  dfs(0, -1);
  if (ok) {
    memset(visited, 0, sizeof(visited));
    dfs(0);
    long long answer = 0LL;
    for (int i = 0; i < n; i++) answer += a[i];
    cout << answer << endl;
  } else {
    cout << "-1" << endl;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  while (t--) solve();
  return 0;
}
