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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
const int maxn = 1e6 + 10, maxm = 2e6 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double PI = acos(-1.0);
int casn, n, m, k;
int num[maxn];
vector<int> g[maxn];
long long aa[maxn];
long long ans[maxn];
int flag;
void dfs(int now, int pre, int dep, long long sum) {
  if (dep % 2 == 0) {
    long long mn = INF;
    for (auto i : g[now]) {
      mn = min(mn, aa[i]);
    }
    if (mn == INF) mn = sum;
    ans[now] = mn - sum;
    sum += ans[now];
    for (auto i : g[now]) {
      dfs(i, now, dep + 1, sum);
    }
  } else {
    ans[now] = aa[now] - sum;
    sum += ans[now];
    for (auto i : g[now]) {
      dfs(i, now, dep + 1, sum);
    }
  }
}
void dfs2(int now, long long sum) {
  sum += ans[now];
  if (aa[now] != -1 && sum != aa[now]) flag = 1;
  for (auto i : g[now]) dfs2(i, sum);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 2; i <= n; ++i) {
    int a;
    cin >> a;
    g[a].push_back(i);
  }
  for (int i = 1; i <= n; ++i) cin >> aa[i];
  ans[1] = aa[1];
  dfs(1, 1, 1, 0);
  for (int i = 1; i <= n; ++i) ans[i] = max(ans[i], 0ll);
  dfs2(1, 0);
  if (flag) {
    cout << -1;
    return 0;
  }
  long long sum = 0;
  for (int i = 1; i <= n; ++i) {
    sum += ans[i];
  }
  cout << sum;
  return 0;
}
