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>
const int N = 1e5 + 5;
const long double pi = 3.141592653;
const long long MOD = 1e9 + 7;
const long long mod = 998244353;
const long long INF = 9223372036854775807LL;
using namespace std;
int dx[9] = {1, -1, 0, 0, 1, 1, -1, -1, 0};
int dy[9] = {0, 0, 1, -1, 1, -1, 1, -1, 0};
vector<long long> g[N], s(N), a(N);
vector<bool> vis(N, false);
void dfs(long long t, long long sum) {
  vis[t] = true;
  if (s[t] != -1)
    a[t] = s[t] - sum;
  else {
    long long val = INF;
    for (long long i = 0; i < (long long)g[t].size(); i++)
      val = min(val, s[g[t][i]]);
    if (val != INF) {
      a[t] = val - sum;
      s[t] = sum + a[t];
    } else {
      s[t] = sum;
      a[t] = 0;
    }
  }
  for (long long i = 0; i < (long long)g[t].size(); i++) {
    if (!vis[g[t][i]]) dfs(g[t][i], sum + a[t]);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  while (T--) {
    long long n, i, x, ans = 0;
    cin >> n;
    for (i = 2; i <= n; i++) {
      cin >> x;
      g[x].push_back(i);
    }
    for (i = 1; i <= n; i++) cin >> s[i];
    dfs(1, 0);
    for (i = 1; i <= n; i++) {
      if (a[i] < 0)
        break;
      else
        ans = ans + a[i];
    }
    if (i <= n)
      cout << -1;
    else
      cout << ans;
  }
  return 0;
}
