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;
namespace IN {
char ss[1 << 17], *A = ss, *B = ss;
inline char gc() {
  if (A == B) {
    B = (A = ss) + fread(ss, 1, 1 << 17, stdin);
    if (A == B) return EOF;
  }
  return *A++;
}
template <typename T>
inline void read(T& x) {
  register char c;
  T y;
  for (c = gc(), x = 0, y = 1; c < 48 || 57 < c; c = gc())
    if (c == '-') y = -1;
  for (; 48 <= c && c <= 57; c = gc()) x = ((x + (x << 2)) << 1) + (c ^ '0');
  x *= y;
}
inline int read() {
  register char c;
  int x, y;
  for (c = gc(), x = 0, y = 1; c < 48 || 57 < c; c = gc())
    if (c == '-') y = -1;
  for (; 48 <= c && c <= 57; c = gc()) x = ((x + (x << 2)) << 1) + (c ^ '0');
  return x * y;
}
template <typename T, typename... Args>
inline void read(T& t, Args&... args) {
  read(t);
  read(args...);
}
template <typename T>
inline void gi(T& x) {
  register char c;
  T y;
  for (c = gc(), x = 0, y = 1; c < 48 || 57 < c; c = gc())
    ;
  for (; 48 <= c && c <= 57; c = gc()) x = ((x + (x << 2)) << 1) + (c ^ '0');
  x *= y;
}
inline int gi() {
  register char c;
  int x, y;
  for (c = gc(), x = 0, y = 1; c < 48 || 57 < c; c = gc())
    ;
  for (; 48 <= c && c <= 57; c = gc()) x = ((x + (x << 2)) << 1) + (c ^ '0');
  return x * y;
}
template <typename T, typename... Args>
inline void gi(T& t, Args&... args) {
  gi(t);
  gi(args...);
}
}  // namespace IN
namespace OUT {
char buf[1 << 21], sta[20];
int p1 = 0, p2 = -1;
inline void flush() { fwrite(buf, 1, p2 + 1, stdout), p2 = -1; }
inline void put(char ch) { buf[++p2] = ch; }
inline void put(string ch) {
  int len = ch.size();
  for (int i = 0; i < len; i++) {
    buf[++p2] = ch[i];
  }
}
inline void print(long long x) {
  if (x < 0) buf[++p2] = 45, x = -x;
  do {
    sta[++p1] = x % 10 + 48;
  } while (x /= 10);
  for (; p1; p1--) buf[++p2] = sta[p1];
}
inline void print(char ch, long long x) {
  if (x < 0) buf[++p2] = 45, x = -x;
  do {
    sta[++p1] = x % 10 + 48;
  } while (x /= 10);
  for (; p1; p1--) buf[++p2] = sta[p1];
  buf[++p2] = ch;
}
template <typename T, typename... Args>
inline void print(char ch, T t, Args... args) {
  print(ch, t);
  print(ch, args...);
}
}  // namespace OUT
using namespace IN;
using namespace OUT;
const int N = 1e5 + 5;
long long n, s[N], fa[N], ans;
int main() {
  gi(n);
  for (long long i = 2; i <= n; ++i) gi(fa[i]);
  for (long long i = 1; i <= n; ++i) read(s[i]);
  for (long long i = 1; i <= n; ++i)
    if (s[i] == -1) s[i] = 0x3f3f3f3f;
  for (long long i = 2; i <= n; ++i) s[fa[i]] = min(s[i], s[fa[i]]);
  for (long long i = 1; i <= n; ++i) {
    if (s[i] < s[fa[i]]) {
      ans = -1;
      break;
    }
    if (s[i] <= 1e9) ans += s[i] - s[fa[i]];
  }
  print(ans), flush();
  return 0;
}
