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;
const long long mod = 1e9 + 7;
const int M = 2e6 + 5;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long expo(long long base, long long pow) {
  long long ans = 1;
  if (pow < 0) return 0;
  while (pow != 0) {
    if (pow & 1 == 1) {
      ans = ans * base;
      ans = ans;
    }
    base *= base;
    pow /= 2;
  }
  return ans;
}
long long expo1(long long base, long long pow) {
  long long ans = 1;
  if (pow < 0) return 0;
  while (pow != 0) {
    if (pow & 1 == 1) {
      ans = ans * base;
      ans = ans % mod;
    }
    base *= base;
    base %= mod;
    pow /= 2;
  }
  return ans;
}
long long inv(long long x) { return expo1(x, mod - 2); }
long long mult_mod(long long a, long long b) {
  a = a % mod;
  b = b % mod;
  long long c = (a * b) % mod;
  return c;
}
long long add_mod(long long a, long long b) {
  a = (a + mod) % mod;
  b = (b + mod) % mod;
  long long c = (a + b) % mod;
  return c;
}
long long fac[(int)(2e6 + 7)], inv_fac[(int)(2e6 + 7)];
void precomp_factorial() {
  fac[0] = 1;
  for (long long i = 1; i <= M; i++) fac[i] = mult_mod(i, fac[i - 1]);
  inv_fac[M] = inv(fac[M]);
  for (long long i = M - 1; i >= 0; i--)
    inv_fac[i] = mult_mod(i + 1, inv_fac[i + 1]);
}
long long nCk(long long n, long long k) {
  if (k == 0) return 1LL;
  long long ans = fac[n];
  ans = mult_mod(ans, inv_fac[n - k]);
  ans = mult_mod(ans, inv_fac[k]);
  return ans;
}
int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};
long long min1[123456], arr[123456], par[123456];
long long sum[123456];
vector<vector<int> > vec;
int cond = 1;
void dfs(int n, int p, int lev) {
  par[n] = p;
  if (lev % 2) {
    arr[n] = 0;
    for (auto x : vec[n]) {
      if (x != p) dfs(x, n, lev + 1);
    }
  } else {
    arr[n] = sum[n] - sum[par[p]];
    for (auto x : vec[n]) {
      if (x != p) dfs(x, n, lev + 1);
    }
  }
}
void dfs1(int n, int p, int lev) {
  if (lev % 2) {
    long long mn = 1e9;
    for (auto x : vec[n]) {
      if (x != p) mn = min(mn, arr[x]);
    }
    if (mn != 1e9)
      arr[n] = mn;
    else
      arr[n] = 0;
    for (auto x : vec[n]) {
      if (x != p) arr[x] -= arr[n];
    }
  }
  for (auto x : vec[n]) {
    if (x != p) dfs1(x, n, lev + 1);
  }
}
int main() {
  int n;
  cin >> n;
  vec.resize(n + 1);
  for (int i = 2; i <= n; i++) {
    int n1;
    cin >> n1;
    vec[i].push_back(n1);
    vec[n1].push_back(i);
  }
  int cnt2 = 0;
  for (int i = 1; i <= n; i++) {
    cin >> sum[i];
  }
  dfs(1, 0, 0);
  dfs1(1, 0, 0);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    if (arr[i] < 0) cond = 0;
    ans += arr[i];
  }
  if (cond)
    cout << ans;
  else
    cout << -1;
}
