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("inline")
using namespace std;
template <class T>
inline T gcd(T x, T y) {
  return y ? gcd(y, x % y) : x;
}
template <class T>
inline T lcm(T x, T y) {
  return x / gcd(x, y) * y;
}
template <class T>
inline void adj(T &x, T y) {
  if (x >= y) x = x % y;
  while (x < 0) x = x + y;
}
template <class T>
inline T qpow(T x, long long y, T z) {
  for (; y; y >>= 1, x = x * x)
    if (y & 1) z = z * x;
  return z;
}
template <class T>
inline T mpow(long long w, T x, long long y, T z) {
  for (; y; y >>= 1, x = x * x, x = x % w)
    if (y & 1) z = z * x, z = z % w;
  return z;
}
template <class T>
inline T exgcd(T a, T b, T &x, T &y) {
  T t = a;
  b ? (t = exgcd(b, a % b, y, x), y = y - (a / b) * x) : (x = 1, y = 0);
  return t;
}
const long long MOD = 1000000007LL;
const int M = 5000005;
const int N = 200005;
int n, fa[N];
long long ans, a[N];
vector<int> g[N];
void dfs(int x) {
  if (a[x] < 0) {
    long long mn = 0x3f3f3f3f3f3f3f3f;
    for (auto &i : g[x]) mn = min(mn, a[i]);
    if (mn == 0x3f3f3f3f3f3f3f3f) mn = a[fa[x]];
    a[x] = mn;
  }
  if (a[x] < a[fa[x]]) {
    puts("-1");
    exit(0);
  }
  for (auto &i : g[x]) dfs(i);
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    scanf("%d", &fa[i]);
    g[fa[i]].push_back(i);
  }
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  dfs(1);
  for (int i = 1; i <= n; i++) ans += a[i] - a[fa[i]];
  printf("%lld\n", ans);
  return 0;
}
