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 inf = 0x3f3f3f3f;
const int maxn = 111111;
vector<int> son[maxn];
int s[maxn];
int d[maxn];
int a[maxn];
bool flag;
void dfs(int u, int sum) {
  if (flag == false) {
    return;
  }
  for (int i = 0; i < son[u].size(); i++) {
    int v = son[u][i];
    d[v] = d[u] + 1;
  }
  if (d[u] % 2 == 0) {
    int mx = inf;
    for (int i = 0; i < son[u].size(); i++) {
      int v = son[u][i];
      mx = min(mx, s[v]);
    }
    if (mx == inf) {
      a[u] = 0;
      s[u] = sum;
    } else if (mx < sum) {
      flag = false;
      return;
    } else {
      a[u] = mx - sum;
      s[u] = mx;
    }
    for (int i = 0; i < son[u].size(); i++) {
      int v = son[u][i];
      dfs(v, s[u]);
    }
  } else {
    a[u] = s[u] - sum;
    for (int i = 0; i < son[u].size(); i++) {
      int v = son[u][i];
      dfs(v, s[u]);
    }
  }
}
int main() {
  int n;
  while (cin >> n) {
    flag = true;
    for (int i = 1; i < n; i++) {
      int fa;
      cin >> fa;
      fa--;
      son[fa].push_back(i);
    }
    for (int i = 0; i < n; i++) {
      cin >> s[i];
    }
    d[0] = 1;
    dfs(0, 0);
    if (flag == false) {
      puts("-1");
    } else {
      long long ans = 0;
      for (int i = 0; i < n; i++) ans += a[i];
      cout << ans << endl;
    }
  }
  return 0;
}
