codeforces 208 E. Blood Cousins （dsu on the tree）

E. Blood Cousins

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Polycarpus got hold of a family relationship tree. The tree describes family relationships of n people, numbered 1 through n. Each person in the tree has no more than one parent.

Let's call person a a 1-ancestor of person b, if a is the parent of b.

Let's call person a a k-ancestor (k > 1) of person b, if person b has a 1-ancestor, and a is a (k - 1)-ancestor of b's 1-ancestor.

Family relationships don't form cycles in the found tree. In other words, there is no person who is his own ancestor, directly or indirectly (that is, who is an x-ancestor for himself, for some x, x > 0).

Let's call two people x and y (x ≠ y) p-th cousins (p > 0), if there is person z, who is a p-ancestor of x and a p-ancestor of y.

Polycarpus wonders how many counsins and what kinds of them everybody has. He took a piece of paper and wrote m pairs of integersvi, pi. Help him to calculate the number of pi-th cousins that person vi has, for each pair vi, pi.

Input

The first input line contains a single integer n (1 ≤ n ≤ 105) — the number of people in the tree. The next line contains n space-separated integers r1, r2, ..., rn, where ri (1 ≤ ri ≤ n) is the number of person i's parent or 0, if person i has no parent. It is guaranteed that family relationships don't form cycles.

The third line contains a single number m (1 ≤ m ≤ 105) — the number of family relationship queries Polycarus has. Next m lines contain pairs of space-separated integers. The i-th line contains numbers vi, pi (1 ≤ vi, pi ≤ n).

Output

Print m space-separated integers — the answers to Polycarpus' queries. Print the answers to the queries in the order, in which the queries occur in the input.

Examples

input

6
0 1 1 0 4 4
7
1 1
1 2
2 1
2 2
4 1
5 1
6 1

output

0 0 1 0 0 1 1 

题目大意：给出一棵家谱树，定义向上走k步到达的节点为该点的k-ancestor.每次询问与v同P-ancestor的节点有多少个。

题解：dsu on the tree

将问题转换成p-ancestor的子树有多少个深度为deep[v]的节点。

#include
#include
#include
#include
#include
#define N 200003
using namespace std;
int tot,point[N],v[N],nxt[N],next[N],head[N],c[N],u[N],mark[N];
int deep[N],size[N],son[N],fa[N][20],mi[20],ans[N],num[N],n,m;
void add(int x,int y)
{
	tot++; nxt[tot]=point[x]; point[x]=tot; v[tot]=y;
	tot++; nxt[tot]=point[y]; point[y]=tot; v[tot]=x;
}
void build(int x,int y,int i)
{
	tot++; next[tot]=head[x]; head[x]=tot; u[tot]=y; c[tot]=i;
}
void solve(int x,int f)
{
	deep[x]=deep[f]+1; size[x]=1;
	for (int i=1;i<=17;i++) {
		if (deep[i]-mi[i]<0) continue;
		fa[x][i]=fa[fa[x][i-1]][i-1];
	}
	for (int i=point[x];i;i=nxt[i]){
		if(v[i]==f) continue;
		fa[v[i]][0]=x;
	    solve(v[i],x);
		size[x]+=size[v[i]];
		if (size[son[x]]>i)&1) x=fa[x][i];
	return x;
}
void change(int x,int f,int val)
{
	num[deep[x]]+=val;
	for (int i=point[x];i;i=nxt[i])
	 if (v[i]!=f&&!mark[v[i]]) change(v[i],x,val);
}
void dfs(int x,int f,bool k)
{
	for (int i=point[x];i;i=nxt[i]) 
	 if (v[i]!=f&&v[i]!=son[x]) dfs(v[i],x,0);
	if (son[x]) dfs(son[x],x,1),mark[son[x]]=1;
	change(x,f,1);
	for (int i=head[x];i;i=next[i])
	 ans[c[i]]=num[u[i]];
	if (son[x]) mark[son[x]]=0;
	if (!k) change(x,f,-1);
}
int main()
{
	freopen("a.in","r",stdin);
	scanf("%d",&n);
	for (int i=1;i<=n;i++) {
		int x; scanf("%d",&x);
		add(x,i);
	}
	solve(0,0);
	scanf("%d",&m); tot=0;
	for (int i=1;i<=m;i++) {
		int x,p; scanf("%d%d",&x,&p);
		if (deep[x]-2