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 int maxn = 1e5 + 100;
int top, n;
int head[maxn], depth[maxn];
long long val[maxn];
struct node {
  int v, next;
} edge[maxn * 2];
inline void add(int u, int v) {
  edge[top].v = v;
  edge[top].next = head[u];
  head[u] = top++;
}
inline void Init() {
  top = 0;
  memset(head, -1, sizeof(head));
}
long long ans;
bool flag;
void dfs(int u, int father) {
  if (val[u] == -1) {
    long long mins = 0x3f3f3f3f;
    for (int i = head[u]; i != -1; i = edge[i].next) {
      int v = edge[i].v;
      if (v != father) {
        mins = min(mins, val[v]);
      }
    }
    val[u] = mins;
  }
  for (int i = head[u]; i != -1; i = edge[i].next) {
    int v = edge[i].v;
    if (v != father) {
      dfs(v, u);
      if (val[v] == 0x3f3f3f3f) {
        val[v] = val[u];
      }
    }
  }
}
void dfs_ans(int u, int father) {
  if (!flag) return;
  for (int i = head[u]; i != -1; i = edge[i].next) {
    int v = edge[i].v;
    if (v != father) {
      long long temp = val[v] - val[u];
      if (temp < 0) {
        flag = false;
        return;
      }
      ans += temp;
      dfs_ans(v, u);
    }
  }
}
int main() {
  cin >> n;
  Init();
  int x;
  for (int i = 1; i <= n - 1; ++i) {
    scanf("%d", &x);
    add(i + 1, x);
    add(x, i + 1);
  }
  for (int i = 1; i <= n; ++i) {
    cin >> val[i];
  }
  ans = val[1];
  flag = true;
  dfs(1, 0);
  dfs_ans(1, 0);
  if (!flag)
    cout << -1 << endl;
  else
    cout << ans << endl;
  return 0;
}
