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 MOD = 1e9 + 7;
typedef struct {
  long long to;
  long long cost;
} edge;
long long N;
long long p[100010];
long long s[100010];
bool ok = true;
vector<long long> tree[100010];
void dfs(long long cur) {
  if (tree[cur].size() == 0) return;
  if (s[cur] == -1) {
    long long mv = (1e18);
    for (auto &e : tree[cur]) {
      dfs(e);
      if (s[e] != -1) {
        mv = min(mv, s[e]);
      }
    }
    if (mv == (1e18)) {
      mv = 0;
    }
    for (auto &e : tree[cur]) {
      if (s[e] == -1) {
        s[e] = 0;
      } else {
        s[e] -= mv;
      }
    }
    s[cur] = mv;
  } else {
    for (auto &e : tree[cur]) {
      dfs(e);
      if (s[e] != -1 && s[cur] > s[e]) {
        ok = false;
        return;
      }
      if (s[e] == -1) {
        s[e] = 0;
      } else {
        s[e] -= s[cur];
      }
    }
  }
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> N;
  for (int i = 0; i < (N - 1); i++) {
    cin >> p[i];
    p[i]--;
    tree[p[i]].push_back(i + 1);
  }
  for (int i = 0; i < (N); i++) {
    cin >> s[i];
  }
  dfs(0);
  long long ans = 0;
  for (int i = 0; i < (N); i++) {
    ans += s[i];
  }
  if (ok) {
    cout << (ans) << endl;
    ;
  } else {
    cout << (-1) << endl;
    ;
  }
}
