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 N = 1e5 + 5, INF = 0x3f3f3f3f;
int n, root = 1;
int s[N], a[N];
int h[N], e[N], ne[N], idx;
void add(int u, int v) { e[idx] = v, ne[idx] = h[u], h[u] = idx++; }
void dfs(int u, int fa, int depth) {
  if (u == root) a[u] = s[u];
  if (!(depth & 1)) {
    int number = INF;
    for (int i = h[u]; ~i; i = ne[i]) {
      int j = e[i];
      number = min(number, s[j]);
    }
    s[u] = number;
  }
  for (int i = h[u]; ~i; i = ne[i]) {
    int j = e[i];
    if (j == fa) continue;
    dfs(j, u, depth + 1);
    if (s[j] != INF) a[j] = s[j] - s[u];
  }
  return;
}
int main() {
  scanf("%d", &n);
  memset(h, -1, sizeof h);
  for (int i = 2; i <= n; i++) {
    int p;
    scanf("%d", &p);
    add(p, i);
  }
  for (int i = 1; i <= n; i++) scanf("%d", s + i);
  dfs(1, -1, 1);
  bool success = true;
  long long res = 0;
  for (int i = 1; i <= n; i++) {
    res += a[i];
    if (a[i] < 0) {
      success = false;
      break;
    }
  }
  if (!success) res = -1;
  printf("%I64d\n", res);
  return 0;
}
