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>
const int maxn = 1e5 + 2e4;
struct node {
  int to, next;
} e[maxn];
int cnt, head[maxn], n, fa[maxn], s[maxn], a[maxn];
void add(int u, int v) {
  e[++cnt].to = v;
  e[cnt].next = head[u];
  head[u] = cnt;
}
void dfs(int x, int sum) {
  if (s[x] <= -1) {
    int mn = 2147483640;
    for (int p = head[x]; p; p = e[p].next) {
      int v = e[p].to;
      mn = std::min(mn, s[v]);
    }
    if (mn == 2147483640) return;
    mn -= sum;
    a[x] = mn;
    sum += a[x];
    for (int p = head[x]; p; p = e[p].next) s[e[p].to] -= sum;
    for (int p = head[x]; p; p = e[p].next) dfs(e[p].to, sum);
    return;
  }
  a[x] = s[x];
  for (int p = head[x]; p; p = e[p].next) {
    int v = e[p].to;
    dfs(v, sum + a[x]);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    int x;
    scanf("%d", &x);
    add(x, i);
    fa[i] = x;
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d", &s[i]);
    if (s[i] != -1 && s[i] < s[1]) {
      printf("-1\n");
      return 0;
    }
  }
  dfs(1, 0);
  long long sum = 0;
  for (int i = 1; i <= n; i++)
    if (a[i] < 0) {
      printf("-1\n");
      return 0;
    } else
      sum += a[i];
  printf("%lld\n", sum);
}
