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 <class T>
inline T lowbit(T x) {
  return x & (-x);
}
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <class T>
inline T Pow(T a, T b, T p) {
  T ret = 1;
  for (; b; b >>= 1, a = a * a % p)
    if (b & 1) (ret *= a) %= p;
  return ret;
}
template <class T>
inline void read(T &ret) {
  T x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  ret = x * f;
}
const int N = 1e5 + 10;
struct edge {
  int v, nxt;
} g[N];
int sz = 0, head[N];
void add(int u, int v) {
  g[++sz] = {v, head[u]};
  head[u] = sz;
}
int n;
long long s[N], ans = 0;
void dfs(int u, int dep, long long sum) {
  if (dep % 2) {
    ans += s[u] - sum;
    sum = s[u];
    for (int i = head[u]; i; i = g[i].nxt) dfs(g[i].v, dep + 1, sum);
  } else {
    long long lim = (long long)0x3f3f3f3f;
    for (int i = head[u]; i; i = g[i].nxt) lim = min(lim, s[g[i].v]);
    if (lim == (long long)0x3f3f3f3f) return;
    if (lim < sum) {
      printf("-1");
      exit(0);
    }
    ans += lim - sum;
    sum = lim;
    for (int i = head[u]; i; i = g[i].nxt) dfs(g[i].v, dep + 1, sum);
  }
}
int main() {
  read(n);
  for (int i = (2); i <= (n); i++) {
    int p;
    read(p);
    add(p, i);
  }
  for (int i = (1); i <= (n); i++) read(s[i]);
  dfs(1, 1, 0);
  printf("%lld\n", ans);
}
