Problem: Snuke has a rooted tree with N vertices, numbered 1 through N. Vertex 1 is the root of the tree, and the parent of Vertex i ( 2\leq i \leq N ) is Vertex P_i ( P_i < i ). There is a number, 0 or 1, written on each vertex. The number written on Vertex i is V_i.

Snuke would like to arrange the vertices of this tree in a horizontal row. Here, for every vertex, there should be no ancestor of that vertex to the right of that vertex.

After arranging the vertices, let X be the sequence obtained by reading the numbers written on the vertices from left to right in the arrangement. Snuke would like to minimize the inversion number of X. Find the minimum possible inversion number of X.

Constraints

* 1 \leq N \leq 2 \times 10^5
* 1 \leq P_i < i ( 2 \leq i \leq N )
* 0 \leq V_i \leq 1 ( 1 \leq i \leq N )
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
P_2 P_3 ... P_N
V_1 V_2 ... V_N


Output

Print the minimum possible inversion number of X.

Examples

Input

6
1 1 2 3 3
0 1 1 0 0 0


Output

4


Input

1

0


Output

0


Input

15
1 2 3 2 5 6 2 2 9 10 1 12 13 12
1 1 1 0 1 1 0 0 1 0 0 1 1 0 0


Output

31

Solution (in CPP):

Solution:
/*
1.树拓扑可以开贪心，子节点合并到父节点
2.若合并前面对后面有影响，可以使用比例排序 
*/
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#define MAXN 200010
using namespace std;
#define int long long
struct Node{int idx,ct1,ct2;};
struct CMP{inline bool operator()(const Node&a,const Node&b){return a.ct1*b.ct2 < a.ct2*b.ct1;}};
int cnt1[MAXN], cnt2[MAXN];
/*
a < b:
count1 a/count2 b < count1 b / count2 b
*/
typedef priority_queue<Node,vector<Node>,CMP> PQC;
PQC q;
int n,fa[MAXN],f[MAXN],u,ans;
int find(int x){return x==f[x]?x:f[x]=find(f[x]);}
signed main(){
	ios_base::sync_with_stdio(false);
	cin>>n;
	for(int i=2;i<=n;++i)cin>>fa[i];
	for(int i=1,t;i<=n;++i){
		cin>>t;
		cnt1[i]=!(cnt2[i]=t);
		f[i]=i;
	}
	for(int i=2;i<=n;++i)
		q.push((Node){i,cnt1[i],cnt2[i]});
	Node t;
	while(!q.empty()){
		u=(t=q.top()).idx;q.pop();
		if(t.ct1==cnt1[u]&&t.ct2==cnt2[u]){//去掉重复的父节点(下面无法像set一样清除父节点) 
			int y=f[u]=find(fa[u]);
			ans+=cnt2[y]*cnt1[u];
			cnt1[y]+=cnt1[u];
			cnt2[y]+=cnt2[u];
			if(y!=1)q.push((Node){y,cnt1[y],cnt2[y]});
		}
	}
	cout<<ans<<endl;
	return 0;
}