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 maxn = 100000 + 7;
const int ooi = 1e9 + 7;
const long long ool = 1e10 + 7;
const int apsz = 26;
const int maxm = 300 + 7;
const int maxb = 19;
const long long mod = 1e9 + 7;
long long s[maxn];
vector<long long> edge[maxn];
long long dep[maxn];
long long vis[maxn];
long long n;
long long a[maxn];
void dfs1(long long u) {
  for (auto v : edge[u]) {
    dep[v] = dep[u] + 1;
    dfs1(v);
  }
}
void dfs2(long long now, long long fa) {
  long long val = -1;
  if (dep[now] % 2 == 0 && edge[now].size() == 0) {
    a[now] = 0;
    return;
  }
  if (dep[now] % 2 == 1) a[now] = s[now] - s[fa];
  for (auto son : edge[now]) {
    if (dep[now] % 2 == 0) {
      long long tmp;
      tmp = s[son] - s[fa];
      if (val == -1 && tmp >= 0) {
        a[now] = tmp;
        s[now] = tmp + s[fa];
        val = tmp;
      } else if (tmp < 0) {
        cout << -1 << endl;
        exit(0);
      } else if (tmp < val) {
        val = tmp;
        a[now] = tmp;
        s[now] = tmp + s[fa];
      }
    }
  }
  for (auto son : edge[now]) {
    dfs2(son, now);
  }
}
void init() {
  cin >> n;
  for (int i = (2); i <= (n); ++i) {
    long long v;
    cin >> v;
    edge[v].push_back(i);
  }
  for (int i = (1); i <= (n); ++i) cin >> s[i];
}
void solve() {
  dep[1] = 1;
  dfs1(1);
  s[0] = 0;
  dfs2(1, 0);
  long long ans = 0;
  for (int i = (1); i <= (n); ++i) ans += a[i];
  cout << ans << endl;
}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  init();
  solve();
  return 0;
}
