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 long long N = 1000006;
using namespace std;
const long long MOD = 1000000007LL;
template <typename T>
T gcd(T a, T b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
template <typename T>
T power(T x, T y, long long m = MOD) {
  T ans = 1;
  while (y > 0) {
    if (y & 1LL) ans = (ans * x) % m;
    y >>= 1LL;
    x = (x * x) % m;
  }
  return ans % m;
}
vector<long long> g[N];
long long s[N], a[N];
void go(long long n, long long p, long long lev) {
  bool leaf = 1;
  for (auto i : g[n]) {
    if (i == p) continue;
    go(i, n, lev + 1);
    leaf = 0;
  }
  if (s[n] == -1) {
    if (leaf) {
      a[n] = 0;
      return;
    }
    long long tv = s[p];
    long long maxi = 1000111000111000111LL;
    for (auto i : g[n]) {
      if (i == p) continue;
      maxi = min(maxi, s[i] - tv);
    }
    a[n] = maxi;
    if (maxi < 0) a[n] = -1000111000111000111LL;
  }
}
void setv(long long n, long long p, long long cs) {
  if (a[n] == -1) {
    a[n] = s[n] - cs;
  }
  for (auto i : g[n]) {
    if (i == p) continue;
    setv(i, n, cs + a[n]);
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  memset(a, -1, sizeof(a));
  for (long long(i) = (0); i < (n - 1); i++) {
    long long x;
    cin >> x;
    --x;
    g[i + 1].emplace_back(x);
    g[x].emplace_back(i + 1);
  }
  for (long long(i) = (0); i < (n); i++) {
    cin >> s[i];
  }
  go(0, -1, 0);
  setv(0, -1, 0);
  long long ans = 0;
  for (long long(i) = (0); i < (n); i++) {
    ans += a[i];
    if (a[i] < 0) {
      cout << "-1";
      return 0;
    }
  }
  cout << ans;
}
