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;
int n;
vector<int> g[100001];
long long s[100001];
int visited[100001] = {0};
int f = 0;
long long ans = 0;
long long depth[100001] = {0};
void dfs1(int x, int d) {
  visited[x] = 1;
  depth[x] = d;
  for (int i = 0; i < g[x].size(); i++) {
    if (visited[g[x][i]] == 0) {
      dfs1(g[x][i], d + 1);
    }
  }
}
void dfs(int x, int val) {
  visited[x] = 1;
  long long minx = 10000000000000;
  if (depth[x] % 2 == 1 && g[x].size() > 1) {
    for (int i = 0; i < g[x].size(); i++) {
      if (depth[g[x][i]] == depth[x] + 1) {
        minx = min(minx, s[g[x][i]]);
      }
    }
    s[x] = minx;
  } else if (s[x] == -1)
    s[x] = val;
  if (s[x] < val) {
    f = 1;
  }
  ans += s[x] - val;
  for (int i = 0; i < g[x].size(); i++) {
    if (visited[g[x][i]] == 0) {
      dfs(g[x][i], s[x]);
    }
  }
}
int main() {
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u;
    cin >> u;
    g[i + 1].push_back(u - 1);
    g[u - 1].push_back(i + 1);
  }
  for (int i = 0; i < n; i++) cin >> s[i];
  dfs1(0, 0);
  for (int i = 0; i < n; i++) visited[i] = 0;
  dfs(0, 0);
  if (f)
    cout << "-1";
  else
    cout << ans;
}
