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;
template <typename tp>
tp read() {
  tp x = 0;
  char c = getchar();
  bool sgn = 0;
  while ((c < '0' || c > '9') && c != '-') c = getchar();
  if (c == '-') sgn = 1, c = getchar();
  while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0', c = getchar();
  return sgn ? -x : x;
}
struct edge {
  long long to, next;
} e[200005 << 1];
int head[200005], cnte;
void add(int u, int v) {
  e[++cnte].to = v;
  e[cnte].next = head[u];
  head[u] = cnte;
}
long long n, a[200005], ans, mn[200005];
void dfs(long long u, long long last) {
  mn[u] = 100000000000000000ll;
  if (a[u] != -1 && a[u] < mn[u]) mn[u] = a[u];
  for (long long i = head[u]; i; i = e[i].next) {
    long long v = e[i].to;
    if (v == last) continue;
    dfs(v, u);
    mn[u] = min(mn[u], mn[v]);
  }
}
void dfs1(long long u, long long last) {
  if (a[u] != -1 && mn[u] < a[u]) {
    cout << -1;
    exit(0);
  }
  for (long long i = head[u]; i; i = e[i].next) {
    long long v = e[i].to;
    if (v == last || mn[v] == 100000000000000000ll) continue;
    ans += mn[v] - mn[u];
    dfs1(v, u);
  }
}
signed main() {
  n = read<long long>();
  for (long long i = 2; i <= n; i++) {
    long long u = read<long long>();
    add(i, u);
    add(u, i);
  }
  for (long long i = 1; i <= n; i++) a[i] = read<long long>();
  dfs(1, 0);
  ans = mn[1];
  dfs1(1, 0);
  cout << ans;
  return 0;
}
