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 N = 2e5 + 5;
const long long mod = 1e9 + 7;
const long long inf = 1e17;
long long dx[] = {0, 1, -1, 0, -1, 1, 1, -1};
long long dy[] = {1, 0, 0, -1, -1, 1, -1, 1};
vector<long long> adj[N], s(N), dp(N);
long long ans;
void dfs(long long start, long long baap, long long h) {
  if (h % 2 == 0) {
    long long mn = 1e17;
    for (auto v : adj[start]) {
      if (v != baap) {
        mn = min(mn, s[v]);
      }
    }
    if (mn != (long long)1e17) {
      dp[start] = max(mn, dp[baap]);
      ans += (dp[start] - dp[baap]);
    }
  } else {
    long long value = s[start] - dp[baap];
    if (value < 0) {
      cout << -1;
      exit(0);
    }
    ans += value;
    dp[start] = s[start];
  }
  for (auto v : adj[start]) {
    if (v != baap) {
      dfs(v, start, h + 1);
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  for (long long i = 2; i <= n; i++) {
    long long par;
    cin >> par;
    adj[i].push_back(par);
    adj[par].push_back(i);
  }
  for (long long i = 1; i <= n; i++) {
    cin >> s[i];
  }
  dfs(1, 0, 1);
  cout << ans;
}
