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:
/*
	直接想贪心
	序列是从祖先到子树里扩展的
	把一个点弄开后，肯定先把所有的 0 给弄进序列里
	这启发我们将点合在一起，即选择了父亲一定再选儿子
	考虑两端序列那个放在前面更优
	发现结果是只要让 cnt_1 / cnt_0 尽量小就行 
*/
#include <bits/stdc++.h>
#define N 200005

using namespace std;

int n;
int fa[N], x[N], y[N]; // x : cnt_1, y : cnt_0
int val[N], o;

namespace UFS {
	int fa[N];
	inline void init() { for(int i = 1; i <= n; ++i) fa[i] = i; }
	inline int find(int u) { return fa[u] == u ? u : fa[u] = find(fa[u]); }
}

struct cmpNode {
	bool operator () (const int &lhs, const int &rhs) {
		if(lhs == 1 || rhs == 1) {
			return lhs > rhs;
		} else if(1LL * x[lhs] * y[rhs] == 1LL * x[rhs] * y[lhs]) {
			return lhs < rhs;
		} else {
			return 1LL * x[lhs] * y[rhs] < 1LL * x[rhs] * y[lhs];
		}
	}
};
set<int, cmpNode> se;

int main() {
	scanf("%d", &n);
	for(int i = 2; i <= n; ++i)
		scanf("%d", &fa[i]);
	for(int i = 1; i <= n; ++i)
		scanf("%d", &val[i]);
	
	UFS::init();
	for(int i = 1; i <= n; ++i) {
		if(val[i]) {
			++x[i];
		} else {
			++y[i];
		}
		se.insert(i);
	}
	
	long long ans = 0;
	set<int, cmpNode>::iterator it;
	for(int i = n; i > 1; --i) {
		it = se.begin();
		int u = *it, v = UFS::find(fa[u]);
		UFS::fa[u] = v;
		se.erase(u), se.erase(v);
		ans += 1LL * y[u] * x[v];
		x[v] += x[u], y[v] += y[u];
		se.insert(v);
	}
	
	printf("%lld\n", ans);
	return 0;
}