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 maxn = 112345;
const int inf = 0x3f3f3f3f;
int n, p[maxn], s[maxn], val[maxn];
bool flag;
vector<int> G[maxn];
void dfs(int u) {
  if (s[u] == -1) {
    int mmax = inf;
    for (int i = 0; i < (int)G[u].size(); ++i)
      mmax = min(mmax, s[G[u][i]] - s[p[u]]);
    if (mmax == inf) mmax = 0;
    if (mmax < 0)
      flag = true;
    else
      val[u] = mmax, s[u] = s[p[u]] + val[u];
  } else
    val[u] = s[u] - s[p[u]];
  for (int i = 0; i < (int)G[u].size(); ++i) {
    int v = G[u][i];
    dfs(v);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; ++i) scanf("%d", &p[i]);
  for (int i = 2; i <= n; ++i) G[p[i]].push_back(i);
  for (int i = 1; i <= n; ++i) scanf("%d", &s[i]);
  dfs(1);
  if (flag)
    puts("-1");
  else {
    long long tot = 0;
    for (int i = 1; i <= n; ++i) tot += val[i];
    cout << tot << endl;
  }
  return 0;
}
