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.

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).

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.

60 1 1 0 4 471 11 22 12 24 15 16 1output
0 0 1 0 0 1 1 
題目大意：給出一棵家譜樹，定義向上走k步到達的節(jié)點為該點的k-ancestor.每次詢問與v同P-ancestor的節(jié)點有多少個。
題解：dsu on the tree
將問題轉(zhuǎn)換成p-ancestor的子樹有多少個深度為deep[v]的節(jié)點。
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#define N 200003using 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]]<size[v[i]]) son[x]=v[i];	}}int get(int x,int k){	for (int i=0;i<=17;i++)	 if ((k>>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<p) {		  ans[i]=1;		  continue;	    }		int f=get(x,p);		build(f,deep[x],i);	}	dfs(0,0,0);	for (int i=1;i<=m;i++) printf("%d ",ans[i]-1);	printf("/n");}