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 = 2e5 + 5;
using namespace std;
vector<int> G[N];
long long b[N];
int vis[N];
long long ans[N];
long long flag = 1;
void dfs(int x, int fa) {
  long long minn = 0x3f3f3f3f;
  if (b[x] == -1) {
    for (int i = 0; i < G[x].size(); i++) {
      int g = G[x][i];
      if (!vis[g]) {
        vis[g] = 1;
        minn = min(minn, b[g]);
        dfs(g, x);
      }
      if (minn != 0x3f3f3f3f)
        b[x] = minn;
      else
        b[x] = b[fa];
    }
  } else {
    for (int i = 0; i < G[x].size(); i++) {
      int g = G[x][i];
      if (!vis[g]) {
        vis[g] = 1;
        dfs(g, x);
      }
    }
  }
}
void dfs2(int x, int fa) {
  for (int i = 0; i < G[x].size(); i++) {
    int g = G[x][i];
    if (!vis[g]) {
      vis[g] = 1;
      ans[g] = b[g] - b[x];
      if (ans[g] < 0) {
        flag = 0;
        return;
      }
      dfs2(g, x);
    }
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 2; i <= n; i++) {
    int u;
    cin >> u;
    G[i].push_back(u);
    G[u].push_back(i);
  }
  for (int i = 1; i <= n; i++) cin >> b[i];
  vis[1] = 1;
  dfs(1, -1);
  memset(vis, 0, sizeof vis);
  vis[1] = 1;
  ans[1] = b[1];
  dfs2(1, -1);
  if (flag) {
    long long res = 0;
    for (int i = 1; i <= n; i++)
      if (ans[i] > 0) res += ans[i];
    cout << res << endl;
  } else
    cout << -1 << endl;
}
