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 N = 1e6 + 10;
const long long INF = 1e10;
const long long inf = -1e18;
const long long MOD = 1e9 + 7;
const long long base = 1000 * 1000 * 1000;
long long n, p[N], a[N], ans, us[N], mn, x;
vector<long long> g[N];
void dfs(long long v, long long last) {
  if (g[v].empty()) {
    return;
  }
  if (a[v] == -1) {
    long long mn = 1e9;
    for (int y : g[v]) {
      mn = min(mn, a[y]);
    }
    if (mn < last) {
      cout << -1 << "\n";
      exit(0);
    }
    ans += mn - last;
    for (int y : g[v]) {
      ans += a[y] - mn;
      dfs(y, last);
    }
  } else {
    for (int y : g[v]) {
      dfs(y, a[v]);
    }
  }
}
void solve() {
  cin >> n;
  for (int i = 2; i <= n; i++) {
    cin >> x;
    g[x].push_back(i);
  }
  for (int i = 1; i <= n; i++) cin >> a[i];
  ans = a[1];
  dfs(1, 0);
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  long long t = 1;
  while (t--) {
    solve();
  }
}
